JEE Trigonometric Functions Questions

Let $$f(x)=2x+3\tan x-\pi$$ defined on $$[-2\pi,2\pi]\setminus\left\{\pm\frac{\pi}{2},\;\pm\frac{3\pi}{2}\right\}$$.

Derivative:
$$f'(x)=2+3\sec^{2}x.$$

Since $$\sec^{2}x\ge 1$$ for every real $$x$$, we have $$f'(x)=2+3\sec^{2}x\gt 0$$. Thus $$f(x)$$ is strictly increasing on every interval in which it is continuous. Consequently, in each such interval $$f(x)=0$$ can have at most one root.

The vertical asymptotes of $$\tan x$$ that lie inside the given domain are at $$x=-\frac{3\pi}{2},\;-\frac{\pi}{2},\;\frac{\pi}{2},\;\frac{3\pi}{2}.$$ These points divide the domain into five continuous intervals:

$$I_1:\;[-2\pi,\,-\frac{3\pi}{2}),\qquad I_2:\;(-\frac{3\pi}{2},\,-\frac{\pi}{2}),$$
$$I_3:\;(-\frac{\pi}{2},\,\frac{\pi}{2}),\qquad I_4:\;(\frac{\pi}{2},\,\frac{3\pi}{2}),\qquad I_5:\;(\frac{3\pi}{2},\,2\pi].$$

$$f(-2\pi)=2(-2\pi)+3\tan(-2\pi)-\pi=-5\pi\lt 0.$$ As $$x\rightarrow\left(-\frac{3\pi}{2}\right)^{-},\;\tan x\rightarrow +\infty\;\Longrightarrow\;f(x)\rightarrow +\infty.$$ With $$f(x)$$ increasing, it must cross zero exactly once in $$I_1$$.

As $$x\rightarrow\left(-\frac{3\pi}{2}\right)^{+},\;\tan x\rightarrow -\infty\;\Longrightarrow\;f(x)\rightarrow -\infty.$$ As $$x\rightarrow\left(-\frac{\pi}{2}\right)^{-},\;\tan x\rightarrow +\infty\;\Longrightarrow\;f(x)\rightarrow +\infty.$$ Again, by monotonicity, exactly one root lies in $$I_2$$.

$$\displaystyle\lim_{x\to(-\frac{\pi}{2})^{+}}f(x)=-\infty,\qquad \lim_{x\to(\frac{\pi}{2})^{-}}f(x)=+\infty.$$ Hence there is exactly one root in $$I_3$$.

$$\displaystyle\lim_{x\to(\frac{\pi}{2})^{+}}f(x)=-\infty,\qquad \lim_{x\to(\frac{3\pi}{2})^{-}}f(x)=+\infty.$$ Therefore one root exists in $$I_4$$.

$$\displaystyle\lim_{x\to(\frac{3\pi}{2})^{+}}f(x)=-\infty.$$ At the right end, $$f(2\pi)=2(2\pi)+3\tan(2\pi)-\pi=4\pi-\pi=3\pi\gt 0.$$ Hence $$f(x)$$ rises from $$-\infty$$ to a positive value, giving one more root in $$I_5$$.

Counting the unique root in each of the five intervals, the total number of solutions in the prescribed domain is $$5$$.

Therefore, the correct choice is Option B.

The given sum contains angles that differ by $$1^\circ$$, so we first convert each term by the identity
$$\frac{1}{\sin A\,\sin B}\;=\;\frac{\cot A-\cot B}{\sin\!\left(B-A\right)}\,.$$ Here $$B-A = 1^\circ$$ for every pair.

Hence each term becomes
$$\frac{1}{\sin k^\circ \,\sin(k+1)^\circ}\;=\;\frac{\cot k^\circ-\cot(k+1)^\circ}{\sin1^\circ}\,.$$ With $$k = 60,62,64,\dots ,118$$ we get

$$\alpha \;=\;\frac{1}{\sin1^\circ}\! \Bigl[\bigl(\cot60^\circ-\cot61^\circ\bigr) +\bigl(\cot62^\circ-\cot63^\circ\bigr) +\cdots +\bigl(\cot118^\circ-\cot119^\circ\bigr)\Bigr].$$

Introduce the compact notation

$$S \;=\;\sum_{k=0}^{29}\Bigl(\cot(60+2k)^\circ-\cot(61+2k)^\circ\Bigr).$$ Then $$\alpha=\dfrac{S}{\sin1^\circ}\,,$$ so our task is to evaluate $$S$$.

Separate even and odd arguments:

$$S =\bigl[\cot60^\circ+\cot62^\circ+\cdots+\cot118^\circ\bigr] -\bigl[\cot61^\circ+\cot63^\circ+\cdots+\cot119^\circ\bigr].$$

Write each angle as either below $$90^\circ$$ or above $$90^\circ$$:

• For $$\theta\lt90^\circ$$, $$\cot\theta^\circ=\tan(90^\circ-\theta^\circ).$$
• For $$\theta=90^\circ+\delta^\circ$$, $$\cot\theta^\circ=-\tan\delta^\circ.$$

Applying these to all terms gives

$$\begin{aligned} S&=\Bigl[\tan30^\circ+\tan28^\circ+\cdots+\tan2^\circ\Bigr] -\Bigl[\tan29^\circ+\tan27^\circ+\cdots+\tan1^\circ\Bigr] \\ &\quad-\Bigl[\tan28^\circ+\tan26^\circ+\cdots+\tan2^\circ\Bigr] +\Bigl[\tan29^\circ+\tan27^\circ+\cdots+\tan1^\circ\Bigr]. \end{aligned}$$

The two long odd-angle sums cancel, and every even-angle term from $$2^\circ$$ to $$28^\circ$$ cancels as well. The only term left is

$$S=\tan30^\circ=\frac{1}{\sqrt3}\,.$$

Therefore

$$\alpha=\frac{S}{\sin1^\circ}=\frac{1}{\sqrt3\,\sin1^\circ}\,.$$

Finally,

$$\left(\frac{\csc1^\circ}{\alpha}\right)^2 =\left(\frac{1/\sin1^\circ}{1/(\sqrt3\,\sin1^\circ)}\right)^2 =\left(\sqrt3\right)^2 =3.$$

Answer: 3

The equation is $$(4 - \sqrt{3})\sin x - 2\sqrt{3}\cos^2 x = \dfrac{-4}{1 + \sqrt{3}}$$.

Rationalizing the RHS: $$\dfrac{-4}{1 + \sqrt{3}} \cdot \dfrac{1 - \sqrt{3}}{1 - \sqrt{3}} = \dfrac{-4(1 - \sqrt{3})}{1 - 3} = \dfrac{-4(1 - \sqrt{3})}{-2} = 2(1 - \sqrt{3}) = 2 - 2\sqrt{3}$$.

Using $$\cos^2 x = 1 - \sin^2 x$$:

$$(4 - \sqrt{3})\sin x - 2\sqrt{3}(1 - \sin^2 x) = 2 - 2\sqrt{3}$$

$$(4 - \sqrt{3})\sin x - 2\sqrt{3} + 2\sqrt{3}\sin^2 x = 2 - 2\sqrt{3}$$

$$2\sqrt{3}\sin^2 x + (4 - \sqrt{3})\sin x - 2 = 0$$

Let $$t = \sin x$$. Using the quadratic formula:

$$t = \dfrac{-(4 - \sqrt{3}) \pm \sqrt{(4 - \sqrt{3})^2 + 16\sqrt{3}}}{4\sqrt{3}}$$

$$(4 - \sqrt{3})^2 = 16 - 8\sqrt{3} + 3 = 19 - 8\sqrt{3}$$

$$19 - 8\sqrt{3} + 16\sqrt{3} = 19 + 8\sqrt{3} = (4 + \sqrt{3})^2$$

So $$t = \dfrac{-(4 - \sqrt{3}) \pm (4 + \sqrt{3})}{4\sqrt{3}}$$

Taking the positive sign: $$t = \dfrac{2\sqrt{3}}{4\sqrt{3}} = \dfrac{1}{2}$$

Taking the negative sign: $$t = \dfrac{-8}{4\sqrt{3}} = \dfrac{-2}{\sqrt{3}} = \dfrac{-2\sqrt{3}}{3}$$. Since $$|\sin x| \le 1$$ and $$2\sqrt{3}/3 \approx 1.155 \gt 1$$, this is rejected.

So $$\sin x = \dfrac{1}{2}$$, giving $$x = \dfrac{\pi}{6} + 2n\pi$$ or $$x = \dfrac{5\pi}{6} + 2n\pi$$.

In $$\left[-2\pi, \dfrac{5\pi}{2}\right]$$:

For $$x = \dfrac{\pi}{6} + 2n\pi$$: $$n = -1$$ gives $$x = -\dfrac{11\pi}{6}$$; $$n = 0$$ gives $$x = \dfrac{\pi}{6}$$; $$n = 1$$ gives $$x = \dfrac{13\pi}{6}$$. All three are in the range.

For $$x = \dfrac{5\pi}{6} + 2n\pi$$: $$n = -1$$ gives $$x = -\dfrac{7\pi}{6}$$; $$n = 0$$ gives $$x = \dfrac{5\pi}{6}$$. Both are in the range. $$n = 1$$ gives $$x = \dfrac{17\pi}{6} \approx 8.9 \gt \dfrac{5\pi}{2} \approx 7.85$$, so excluded.

Total number of solutions = 5.

Hence, the correct answer is Option D.

We are given the system of equations:

$$2\sin^{2}\theta = \cos 2\theta \quad \text{(1)}$$

$$2\cos^{2}\theta = 3\sin\theta \quad \text{(2)}$$

We need to find all $$\theta \in [0, 2\pi]$$ satisfying both equations and sum them.

First, recall the double-angle identity for cosine: $$\cos 2\theta = 1 - 2\sin^{2}\theta$$.

Substitute this into equation (1):

$$2\sin^{2}\theta = 1 - 2\sin^{2}\theta$$

Bring all terms to one side:

$$2\sin^{2}\theta - 1 + 2\sin^{2}\theta = 0$$

$$4\sin^{2}\theta - 1 = 0$$

$$4\sin^{2}\theta = 1$$

$$\sin^{2}\theta = \frac{1}{4}$$

Thus, $$\sin\theta = \pm \frac{1}{2}$$.

Now, from equation (2): $$2\cos^{2}\theta = 3\sin\theta$$.

Since, the LHS of equation 2 will always be positive, the only value for the which the RHS is positive is $$\sin\theta=\frac{1}{2}$$.

Now, find all $$\theta \in [0, 2\pi]$$ such that $$\sin\theta = \frac{1}{2}$$.

Sine is positive in the first and second quadrants. The reference angle is $$\frac{\pi}{6}$$.

In first quadrant: $$\theta = \frac{\pi}{6}$$.

In second quadrant: $$\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$.

Verify both solutions satisfy the original equations.

For $$\theta = \frac{\pi}{6}$$:

Equation (1): $$2\sin^{2}\left(\frac{\pi}{6}\right) = 2 \left(\frac{1}{2}\right)^2 = 2 \times \frac{1}{4} = \frac{1}{2}$$, $$\cos 2\left(\frac{\pi}{6}\right) = \cos \frac{\pi}{3} = \frac{1}{2}$$, so equal.

Equation (2): $$2\cos^{2}\left(\frac{\pi}{6}\right) = 2 \left(\frac{\sqrt{3}}{2}\right)^2 = 2 \times \frac{3}{4} = \frac{3}{2}$$, $$3\sin\left(\frac{\pi}{6}\right) = 3 \times \frac{1}{2} = \frac{3}{2}$$, so equal.

For $$\theta = \frac{5\pi}{6}$$:

Equation (1): $$2\sin^{2}\left(\frac{5\pi}{6}\right) = 2 \left(\frac{1}{2}\right)^2 = \frac{1}{2}$$, $$\cos 2\left(\frac{5\pi}{6}\right) = \cos \frac{5\pi}{3} = \cos \left(2\pi - \frac{\pi}{3}\right) = \cos \left(-\frac{\pi}{3}\right) = \cos \frac{\pi}{3} = \frac{1}{2}$$, so equal.

Equation (2): $$2\cos^{2}\left(\frac{5\pi}{6}\right) = 2 \left(-\frac{\sqrt{3}}{2}\right)^2 = 2 \times \frac{3}{4} = \frac{3}{2}$$, $$3\sin\left(\frac{5\pi}{6}\right) = 3 \times \frac{1}{2} = \frac{3}{2}$$, so equal.

No other solutions exist in $$[0, 2\pi]$$ as the value of sine is negative in $$[\pi,2\pi]$$.

Sum of solutions: $$\frac{\pi}{6} + \frac{5\pi}{6} = \frac{6\pi}{6} = \pi$$.

The sum is $$\pi$$, which corresponds to option C.

We need to find the value of $$(\sin 70°)(\cot 10° \cot 70° - 1)$$.

$$\sin 70° \left(\frac{\cos 10° \cos 70°}{\sin 10° \sin 70°} - 1\right) = \sin 70° \cdot \frac{\cos 10° \cos 70° - \sin 10° \sin 70°}{\sin 10° \sin 70°}$$

$$\cos 10° \cos 70° - \sin 10° \sin 70° = \cos(10° + 70°) = \cos 80°$$.

$$= \sin 70° \cdot \frac{\cos 80°}{\sin 10° \sin 70°} = \frac{\cos 80°}{\sin 10°}$$

Since $$\cos 80° = \sin 10°$$:

$$= \frac{\sin 10°}{\sin 10°} = 1$$

The answer is $$\boxed{1}$$, which corresponds to Option 2.

Let the direction angles of the line with the positive $$x$$-, $$y$$- and $$z$$-axes be $$\alpha , \beta , \gamma$$, respectively.
Their direction cosines are

$$l = \cos\alpha,\; m = \cos\beta,\; n = \cos\gamma$$

For every line in space, the direction cosines satisfy the identity

$$l^{2}+m^{2}+n^{2}=1 \;$$ $$-(1)$$

According to the question,

$$\beta = \dfrac{\alpha}{2},\quad \gamma = \dfrac{\alpha}{2}$$ $$-(2)$$

Substituting $$l,m,n$$ and relation $$(2)$$ into $$(1)$$ gives

$$\cos^{2}\alpha + \cos^{2}\!\left(\dfrac{\alpha}{2}\right) + \cos^{2}\!\left(\dfrac{\alpha}{2}\right) = 1$$

$$\Longrightarrow \cos^{2}\alpha + 2\cos^{2}\!\left(\dfrac{\alpha}{2}\right) = 1$$ $$-(3)$$

Use the half-angle identity $$\cos^{2}\!\left(\dfrac{\alpha}{2}\right)=\dfrac{1+\cos\alpha}{2}$$.
Substituting in $$(3)$$:

$$\cos^{2}\alpha + 2\left( \dfrac{1+\cos\alpha}{2} \right) = 1$$

$$\Longrightarrow \cos^{2}\alpha + 1 + \cos\alpha = 1$$

$$\Longrightarrow \cos^{2}\alpha + \cos\alpha = 0$$

Factorising:

$$\cos\alpha\,(\cos\alpha + 1) = 0$$

Therefore

$$\cos\alpha = 0 \quad \text{or} \quad \cos\alpha = -1$$

The direction angles lie in $$0 \le \alpha \le \pi$$.

If $$\cos\alpha = 0$$, then $$\alpha = \dfrac{\pi}{2}$$.
Using $$(2)$$, $$\beta = \dfrac{\alpha}{2} = \dfrac{\pi}{4}$$.

If $$\cos\alpha = -1$$, then $$\alpha = \pi$$.
Using $$(2)$$, $$\beta = \dfrac{\alpha}{2} = \dfrac{\pi}{2}$$.

Thus the possible values of $$\beta$$ are $$\dfrac{\pi}{4}$$ and $$\dfrac{\pi}{2}$$.

Required sum

$$\beta_{\text{sum}} = \dfrac{\pi}{4} + \dfrac{\pi}{2} = \dfrac{3\pi}{4}$$

Hence the sum of all possible values of $$\beta$$ is $$\dfrac{3\pi}{4}$$, corresponding to Option A.

Let $$s = \sin^2\theta$$ and $$c = \cos^2\theta$$. Then $$s + c = 1$$ since $$\sin^2\theta + \cos^2\theta = 1$$.

The given equation is $$10\sin^4\theta + 15\cos^4\theta = 6$$. Re-write it using $$s$$ and $$c$$:
$$10s^2 + 15c^2 = 6$$.

Because $$c = 1 - s$$, substitute to get a quadratic in $$s$$:
$$10s^2 + 15(1 - s)^2 = 6$$

Simplify:
$$10s^2 + 15(1 - 2s + s^2) = 6$$
$$10s^2 + 15 - 30s + 15s^2 = 6$$
$$25s^2 - 30s + 15 - 6 = 0$$
$$25s^2 - 30s + 9 = 0$$ $$-(1)$$

Compute the discriminant of $$(1)$$:
$$\Delta = (-30)^2 - 4 \times 25 \times 9 = 900 - 900 = 0$$.

Since $$\Delta = 0$$, there is one repeated root:
$$s = \frac{30}{2 \times 25} = \frac{3}{5}$$.

Therefore
$$\sin^2\theta = s = \frac{3}{5}\quad\text{and}\quad \cos^2\theta = 1 - s = \frac{2}{5}$$.

Compute the higher-power reciprocals needed.
$$\csc^6\theta = \frac{1}{\sin^6\theta} = \frac{1}{\left(\dfrac{3}{5}\right)^3} = \frac{1}{\dfrac{27}{125}} = \frac{125}{27}$$
$$\sec^6\theta = \frac{1}{\cos^6\theta} = \frac{1}{\left(\dfrac{2}{5}\right)^3} = \frac{1}{\dfrac{8}{125}} = \frac{125}{8}$$
$$\sec^8\theta = \frac{1}{\cos^8\theta} = \frac{1}{\left(\dfrac{2}{5}\right)^4} = \frac{1}{\dfrac{16}{625}} = \frac{625}{16}$$

Form the required numerator:
$$27\csc^6\theta + 8\sec^6\theta = 27\left(\frac{125}{27}\right) + 8\left(\frac{125}{8}\right) = 125 + 125 = 250$$

Form the required denominator:
$$16\sec^8\theta = 16\left(\frac{625}{16}\right) = 625$$

Hence
$$\dfrac{27\csc^6\theta + 8\sec^6\theta}{16\sec^8\theta} = \dfrac{250}{625} = \dfrac{2}{5}$$.

The value equals $$\dfrac{2}{5}$$, which matches Option A.

Given: $$\sin x + \sin^2 x = 1$$, which means $$\sin x = 1 - \sin^2 x = \cos^2 x$$.

Also: $$\tan x = \frac{\sin x}{\cos x}$$, and since $$\sin x = \cos^2 x$$:

$$\tan^2 x = \frac{\sin^2 x}{\cos^2 x} = \frac{\sin^2 x}{\sin x} = \sin x = \cos^2 x$$

So $$\cos^2 x = \tan^2 x$$. Let $$t = \cos^2 x = \tan^2 x$$.

The expression becomes:

$$ (t^6 + t^6) + 3(t^5 + t^5 + t^4 + t^4) + (t^3 + t^3) $$

$$ = 2t^6 + 6t^5 + 6t^4 + 2t^3 = 2t^3(t^3 + 3t^2 + 3t + 1) = 2t^3(t+1)^3 $$

Now we need to find $$t$$. From $$\sin x = \cos^2 x$$ and $$\sin^2 x + \cos^2 x = 1$$:

$$\sin^2 x + \sin x = 1$$ (which is the given condition).

$$t = \cos^2 x = \sin x$$, and $$t + t^2 = 1$$, so $$t^2 + t = 1$$, i.e., $$t + 1 = \frac{1}{t}$$... actually:

$$t^2 + t - 1 = 0 \Rightarrow t = \frac{-1+\sqrt{5}}{2}$$ (taking positive root since $$x \in (0, \pi/2)$$).

So $$t(t+1) = t \cdot \frac{1}{t} \cdot t$$... Let me recalculate:

From $$t^2 + t = 1$$: $$t(t+1) = 1$$.

Therefore: $$2t^3(t+1)^3 = 2[t(t+1)]^3 = 2(1)^3 = 2$$.

The correct answer is Option 4: 2.

We have to evaluate the finite sum

$$S=\displaystyle\sum_{r=1}^{13}\frac{1}{\sin\Bigl(\tfrac{\pi}{4}+(r-1)\tfrac{\pi}{6}\Bigr)\; \sin\Bigl(\tfrac{\pi}{4}+r\tfrac{\pi}{6}\Bigr)}.$$

Put $$A_r=\tfrac{\pi}{4}+(r-1)\tfrac{\pi}{6}\qquad (r=1,2,\dots,13).$$
Then the general term of the series can be rewritten as

$$\frac{1}{\sin A_r\;\sin(A_r+\tfrac{\pi}{6})}.$$

Step 1  : Express the reciprocal product through cotangents

The identity $$\cot x-\cot y=\dfrac{\sin(y-x)}{\sin x\;\sin y}$$ holds for all $$x,y$$ (provided the sines are non-zero). Putting $$x=A_r,\;y=A_r+\tfrac{\pi}{6}$$ gives

$$\frac{1}{\sin A_r\;\sin(A_r+\tfrac{\pi}{6})}= \frac{\cot A_r-\cot(A_r+\tfrac{\pi}{6})}{\sin\bigl((A_r+\tfrac{\pi}{6})-A_r\bigr)} =\frac{\cot A_r-\cot(A_r+\tfrac{\pi}{6})}{\sin(\tfrac{\pi}{6})}.$$

Since $$\sin(\tfrac{\pi}{6})=\tfrac12,$$ we obtain the compact form

$$\frac{1}{\sin A_r\;\sin(A_r+\tfrac{\pi}{6})}=2\bigl[\cot A_r-\cot(A_r+\tfrac{\pi}{6})\bigr].$$

Step 2  : Substitute and observe telescoping

Using the above identity, the required sum becomes

$$\begin{aligned} S&=2\sum_{r=1}^{13}\Bigl[\cot A_r-\cot(A_r+\tfrac{\pi}{6})\Bigr]\\ &=2\Bigl[(\cot A_1-\cot A_2)+(\cot A_2-\cot A_3)+\dots+(\cot A_{13}-\cot A_{14})\Bigr]. \end{aligned}$$

All the interior cotangents cancel in pairs, leaving only the first and the last terms:

$$S=2\bigl[\cot A_1-\cot A_{14}\bigr].$$

Step 3  : Evaluate the boundary cotangents

Compute the angles:

$$A_1=\tfrac{\pi}{4},\qquad A_{14}=A_1+13\cdot\tfrac{\pi}{6}=\tfrac{\pi}{4}+\tfrac{13\pi}{6} =\tfrac{29\pi}{12}.$$

The cotangents are

$$\cot A_1=\cot\!\bigl(\tfrac{\pi}{4}\bigr)=1,$$

and, using the $$\pi$$-periodicity of cotangent,

$$\cot A_{14}=\cot\!\bigl(\tfrac{29\pi}{12}\bigr) =\cot\!\bigl(\tfrac{29\pi}{12}-2\pi\bigr) =\cot\!\bigl(\tfrac{5\pi}{12}\bigr)=\cot 75^{\circ}.$$

Because $$\cot 75^{\circ}=\tan 15^{\circ}$$ and $$\tan 15^{\circ}=2-\sqrt3,$$ we have

$$\cot A_{14}=2-\sqrt3.$$

Step 4  : Obtain the numerical value of $$S$$

$$\begin{aligned} S&=2\bigl[\;1-(2-\sqrt3)\bigr]\\ &=2\bigl[-1+\sqrt3\bigr]\\ &=2\sqrt3-2. \end{aligned}$$

Step 5  : Identify $$a$$ and $$b$$

The sum is of the form $$a\sqrt3+b$$ with integers $$a=2,\;b=-2.$$ Therefore

$$a^{2}+b^{2}=2^{2}+(-2)^{2}=4+4=8.$$

Hence the required value is $$\boxed{8}$$, which corresponds to Option D.

We need to solve $$2\sqrt{2}\cos^2\theta + (2 - \sqrt{6})\cos\theta - \sqrt{3} = 0$$ for $$\theta \in [-2\pi, 2\pi]$$.

Let $$u = \cos\theta$$. The equation becomes $$2\sqrt{2}u^2 + (2 - \sqrt{6})u - \sqrt{3} = 0$$.

Using the quadratic formula: $$u = \frac{-(2-\sqrt{6}) \pm \sqrt{(2-\sqrt{6})^2 + 4 \cdot 2\sqrt{2} \cdot \sqrt{3}}}{2 \cdot 2\sqrt{2}}$$.

Computing the discriminant: $$(2-\sqrt{6})^2 + 8\sqrt{6} = 4 - 4\sqrt{6} + 6 + 8\sqrt{6} = 10 + 4\sqrt{6} = (2 + \sqrt{6})^2$$.

So $$u = \frac{(\sqrt{6}-2) \pm (2+\sqrt{6})}{4\sqrt{2}}$$.

Case 1: $$u = \frac{\sqrt{6}-2+2+\sqrt{6}}{4\sqrt{2}} = \frac{2\sqrt{6}}{4\sqrt{2}} = \frac{\sqrt{6}}{2\sqrt{2}} = \frac{\sqrt{3}}{2}$$.

Case 2: $$u = \frac{\sqrt{6}-2-2-\sqrt{6}}{4\sqrt{2}} = \frac{-4}{4\sqrt{2}} = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$$.

For $$\cos\theta = \frac{\sqrt{3}}{2}$$: $$\theta = \pm\frac{\pi}{6}$$. In $$[-2\pi, 2\pi]$$, the solutions are $$\theta = \frac{\pi}{6}, -\frac{\pi}{6}, 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}, -(2\pi - \frac{\pi}{6}) = -\frac{11\pi}{6}$$. That gives 4 solutions.

For $$\cos\theta = -\frac{\sqrt{2}}{2}$$: $$\theta = \pm\frac{3\pi}{4}$$. In $$[-2\pi, 2\pi]$$, the solutions are $$\theta = \frac{3\pi}{4}, -\frac{3\pi}{4}, 2\pi - \frac{3\pi}{4} = \frac{5\pi}{4}, -(2\pi - \frac{3\pi}{4}) = -\frac{5\pi}{4}$$. That gives 4 solutions.

Total number of solutions = $$4 + 4 = 8$$.

Hence, the correct answer is Option C.

Rewrite the equation in a single trigonometric variable.
Let $$x=\csc\theta$$. Then the given equation becomes

$$\sqrt3\,x^{2}-2(\sqrt3-1)\,x-4=0\qquad -(1)$$

Solve the quadratic for $$x$$.
For a quadratic $$ax^{2}+bx+c=0$$ the roots are $$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$.
Here $$a=\sqrt3,\; b=-2(\sqrt3-1)=-2\sqrt3+2,\; c=-4$$.

Compute the discriminant:

$$D=b^{2}-4ac$$
$$=( -2\sqrt3+2 )^{2}-4(\sqrt3)(-4)$$
$$=16-8\sqrt3+16\sqrt3$$
$$=16+8\sqrt3\;.\qquad -(2)$$

Simplify $$\sqrt{D}$$ by expressing it as $$\sqrt{m}+\sqrt{n}$$:

Find $$m,n$$ such that $$m+n=16$$ and $$2\sqrt{mn}=8\sqrt3\Rightarrow mn=48$$.
The numbers satisfying these are $$m=12,\;n=4$$.
Hence $$\sqrt D=\sqrt{12}+\sqrt4=2\sqrt3+2\;.\qquad -(3)$$

Substitute into the quadratic-formula expression:

$$x=\frac{-b\pm\sqrt D}{2a}=\frac{2\sqrt3-2\;\pm\;(2\sqrt3+2)}{2\sqrt3}$$

Case 1: with $$+$$ sign

$$x=\frac{2\sqrt3-2+2\sqrt3+2}{2\sqrt3}=\frac{4\sqrt3}{2\sqrt3}=2$$

Case 2: with $$-$$ sign

$$x=\frac{2\sqrt3-2-(2\sqrt3+2)}{2\sqrt3}=\frac{-4}{2\sqrt3}=-\frac{2}{\sqrt3}=-\frac{2\sqrt3}{3}$$

Therefore the two admissible csc-values are

$$\csc\theta=2\qquad\text{or}\qquad\csc\theta=-\frac{2\sqrt3}{3}\;.\qquad -(4)$$

Convert each value to a sine condition (remember $$\csc\theta=\frac1{\sin\theta}$$):

$$\csc\theta=2\;\Longrightarrow\;\sin\theta=\frac12$$
$$\csc\theta=-\frac{2\sqrt3}{3}\;\Longrightarrow\;\sin\theta=-\frac{\sqrt3}{2}\qquad -(5)$$

Now list all angles in the interval $$\left[-\frac{7\pi}{6},\,\frac{4\pi}{3}\right]$$ (i.e. from $$-210^{\circ}$$ to $$240^{\circ}$$) satisfying each sine value, taking care to keep the interval end-points because sine is non-zero there.

For $$\sin\theta=\frac12$$:
Standard positions: $$\theta=\frac{\pi}{6},\; \frac{5\pi}{6}$$.
Adding or subtracting integral multiples of $$2\pi$$ gives other coterminal angles. Within the required interval we obtain

$$\theta=-\frac{7\pi}{6},\; \frac{\pi}{6},\; \frac{5\pi}{6}$$

Thus, 3 solutions from this case.

For $$\sin\theta=-\frac{\sqrt3}{2}$$:
Standard positions: $$\theta=\frac{4\pi}{3},\; \frac{5\pi}{3}$$.
Again using periodicity and checking the interval boundaries, the admissible angles are

$$\theta=-\frac{2\pi}{3},\; -\frac{\pi}{3},\; \frac{4\pi}{3}$$

Thus, 3 solutions from this case as well.

Add the counts from both cases:

Total number of solutions $$=3+3=6$$

Hence the required number of solutions is $$6$$, which corresponds to Option A.

The given equation is
$$\cos 2\theta \,\cos\frac{\theta}{2}+\cos\frac{5\theta}{2}=2\cos^{3}\frac{5\theta}{2},\qquad \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].$$

Put $$t=\frac{\theta}{2}.$$
Then $$\theta=2t$$ and the interval becomes $$t\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right].$$
The equation turns into
$$\cos 4t\;\cos t+\cos 5t=2\cos^{3}5t \quad -(1).$$

Use the product-to-sum identity $$\cos A\cos B=\tfrac{1}{2}\bigl[\cos(A+B)+\cos(A-B)\bigr]$$ on the first term:
$$\cos 4t\;\cos t=\tfrac{1}{2}\bigl[\cos(4t+t)+\cos(4t-t)\bigr] =\tfrac{1}{2}\bigl[\cos 5t+\cos 3t\bigr].$$

Substitute this in $$(1):$$
$$\tfrac{1}{2}\bigl[\cos 5t+\cos 3t\bigr]+\cos 5t =2\cos^{3}5t.$$ Combine like terms:
$$(\tfrac{1}{2}+1)\cos 5t+\tfrac{1}{2}\cos 3t =2\cos^{3}5t,$$ $$\tfrac{3}{2}\cos 5t+\tfrac{1}{2}\cos 3t =2\cos^{3}5t.$$ Multiply by $$2:$$ $$3\cos 5t+\cos 3t=4\cos^{3}5t \quad -(2).$$

Recall the triple-angle identity $$\cos 3x=4\cos^{3}x-3\cos x,$$ which can be rearranged as
$$4\cos^{3}x=\cos 3x+3\cos x.$$ Putting $$x=5t$$ gives
$$4\cos^{3}5t=\cos 15t+3\cos 5t.$$ Replace the right side of $$(2)$$ with this expression:
$$3\cos 5t+\cos 3t=\cos 15t+3\cos 5t.$$ The terms $$3\cos 5t$$ cancel, leaving the simple equation
$$\cos 3t=\cos 15t \quad -(3).$$

For $$\cos\alpha=\cos\beta,$$ the solutions are
$$\alpha=2k\pi\pm\beta,\qquad k\in\mathbb{Z}.$$ Apply this to $$(3)$$:

$$3t=2k\pi+15t \;\Longrightarrow\; -12t=2k\pi \;\Longrightarrow\; t=-\frac{k\pi}{6}.$$ With $$t\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right],$$ choose integers $$k$$ so that $$-\frac{k\pi}{6}$$ lies in the interval.
  • $$k=0:\;t=0$$
  • $$k=1:\;t=-\frac{\pi}{6}$$
  • $$k=-1:\;t=\frac{\pi}{6}$$
Further $$|k|\ge 2$$ gives $$|t|\gt\pi/4,$$ so they are rejected.
Thus Case 1 contributes $$t=-\frac{\pi}{6},\,0,\,\frac{\pi}{6}.$$

$$3t=2k\pi-15t \;\Longrightarrow\;18t=2k\pi \;\Longrightarrow\; t=\frac{k\pi}{9}.$$ Again keep the values inside $$\left[-\frac{\pi}{4},\frac{\pi}{4}\right]:$$
  • $$k=0:\;t=0$$ (already counted)
  • $$k=\pm1:\;t=\pm\frac{\pi}{9}$$
  • $$k=\pm2:\;t=\pm\frac{2\pi}{9}$$
  • $$|k|\ge3$$ violates the interval.
Case 2 contributes $$t=-\frac{2\pi}{9},\,-\frac{\pi}{9},\,0,\,\frac{\pi}{9},\,\frac{2\pi}{9}.$$

Collecting distinct solutions of $$t$$ inside the interval:
$$t=-\frac{2\pi}{9},\;-\frac{\pi}{6},\;-\frac{\pi}{9},\;0,\;\frac{\pi}{9},\;\frac{\pi}{6},\;\frac{2\pi}{9}.$$ Thus there are $$7$$ valid $$t$$ values.

Since $$\theta=2t,$$ each $$t$$ gives a unique $$\theta$$ in the original interval $$\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$, so the number of solutions for $$\theta$$ is also $$7$$.

Hence, the correct option is Option A (7).

We are given $$\frac{\pi}{2}\lt x\lt \pi$$ and $$\cot x=\frac{-5}{\sqrt{11}}.$$
Since $$x$$ lies in the second quadrant, $$\sin x\gt 0$$ and $$\cos x\lt 0.$$ Write $$\cot x=\dfrac{\cos x}{\sin x}=\dfrac{-5}{\sqrt{11}}.$$ Choose a common factor $$k$$ such that

$$\cos x=-5k,\qquad\sin x=\sqrt{11}\,k.$$ Using $$\sin^2x+\cos^2x=1$$ gives $$(-5k)^2+(\sqrt{11}\,k)^2=1\;\Longrightarrow\;25k^2+11k^2=36k^2=1\;\Longrightarrow\;k=\frac16.$$

Thus $$\sin x=\frac{\sqrt{11}}6,\qquad\cos x=-\frac56.$$

------------------------------------------------------------

1. Half-angle values

Because $$\dfrac{\pi}{4}\lt\dfrac{x}{2}\lt\dfrac{\pi}{2}$$ (first quadrant), both $$\sin\dfrac{x}{2}$$ and $$\cos\dfrac{x}{2}$$ are positive.

Using the half-angle identities:

$$\cos x=1-2\sin^2\frac{x}{2} \;\Longrightarrow\; -\frac56=1-2\sin^2\frac{x}{2}$$ $$\;\Longrightarrow\; 2\sin^2\frac{x}{2}=\frac{11}{6} \;\Longrightarrow\; \sin^2\frac{x}{2}=\frac{11}{12} \;\Longrightarrow\; \sin\frac{x}{2}=\frac{\sqrt{11}}{2\sqrt{3}}.$$

Then $$\cos\frac{x}{2}=\sqrt{1-\sin^2\frac{x}{2}} =\sqrt{1-\frac{11}{12}} =\sqrt{\frac1{12}} =\frac1{2\sqrt{3}}.$$

------------------------------------------------------------

2. Simplifying the required expression

Let $$A=\frac{11x}{2},\qquad E=\left(\sin\frac{11x}{2}\right)(\sin6x-\cos6x)+\left(\cos\frac{11x}{2}\right)(\sin6x+\cos6x).$$ Rewrite with $$A$$:

$$E=\sin A(\sin6x-\cos6x)+\cos A(\sin6x+\cos6x).$$

Separate the $$\sin6x$$ and $$\cos6x$$ terms:

$$E=\sin6x(\sin A+\cos A)+\cos6x(\cos A-\sin A).$$

Use the identities $$\sin A+\cos A=\sqrt2\,\sin\!\left(A+\frac{\pi}{4}\right),\qquad \cos A-\sin A=\sqrt2\,\cos\!\left(A+\frac{\pi}{4}\right).$$

Hence

$$E =\sqrt2\Bigl[\sin6x\sin\!\left(A+\frac{\pi}{4}\right)+\cos6x\cos\!\left(A+\frac{\pi}{4}\right)\Bigr] =\sqrt2\,\cos\!\left[6x-\left(A+\frac{\pi}{4}\right)\right],$$

because $$\cos(\theta-\phi)=\cos\theta\cos\phi+\sin\theta\sin\phi.$$ Substitute $$A=\dfrac{11x}{2}$$:

$$6x-\left(\frac{11x}{2}+\frac{\pi}{4}\right) =\frac{12x-11x}{2}-\frac{\pi}{4} =\frac{x}{2}-\frac{\pi}{4}.$$

Therefore

$$E=\sqrt2\,\cos\!\left(\frac{x}{2}-\frac{\pi}{4}\right).$$

------------------------------------------------------------

3. Evaluating $$\displaystyle\cos\!\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)$$

Using $$\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$$ with $$\alpha=\frac{x}{2},\;\beta=\frac{\pi}{4},$$

$$\cos\!\left(\frac{x}{2}-\frac{\pi}{4}\right) =\cos\frac{x}{2}\cos\frac{\pi}{4}+\sin\frac{x}{2}\sin\frac{\pi}{4} =\cos\frac{x}{2}\cdot\frac{\sqrt2}{2}+\sin\frac{x}{2}\cdot\frac{\sqrt2}{2}$$ $$=\frac{\sqrt2}{2}\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right) =\frac{\sqrt2}{2}\left(\frac1{2\sqrt3}+\frac{\sqrt{11}}{2\sqrt3}\right) =\frac{\sqrt2}{2}\,\frac{1+\sqrt{11}}{2\sqrt3} =\frac{\sqrt2(1+\sqrt{11})}{4\sqrt3}.$$

------------------------------------------------------------

4. Final result

$$E=\sqrt2\;\times\;\frac{\sqrt2(1+\sqrt{11})}{4\sqrt3} =\frac{2(1+\sqrt{11})}{4\sqrt3} =\frac{1+\sqrt{11}}{2\sqrt3}.$$

Hence the required value is $$\boxed{\dfrac{\sqrt{11}+1}{2\sqrt{3}}}.$$

Option B is correct.

Let $$\alpha = \arcsin\!\left(\dfrac{2}{3}\right)$$. Numerically $$\alpha \approx 0.7297\text{ rad }(41.81^{\circ})$$.

Step 1 : Reduce the given trigonometric equation
Given $$2\sin^{3}x+\sin 2x \cos x+4\sin x-4=0$$.

Use $$\sin 2x = 2\sin x \cos x$$ and $$\sin^{2}x = 1-\cos^{2}x$$:

$$2\sin^{3}x = 2\sin x(1-\cos^{2}x)=2\sin x-2\sin x\cos^{2}x$$

$$\sin 2x \cos x = 2\sin x\cos^{2}x$$

Adding these inside the equation, the $$\pm2\sin x\cos^{2}x$$ terms cancel:

$$\bigl(2\sin x-2\sin x\cos^{2}x\bigr)+2\sin x\cos^{2}x+4\sin x-4=0$$

$$6\sin x-4=0 \;\;\Longrightarrow\;\; \sin x=\dfrac{2}{3}$$

Step 2 : Count the number of solutions of $$\sin x=\dfrac{2}{3}$$ in $$[0,\;n\dfrac{\pi}{2}]$$

General solutions of $$\sin x=\dfrac{2}{3}$$ are

$$x = \alpha + 2k\pi \quad\text{and}\quad x = \pi-\alpha+2k\pi,\;k\in\mathbb{Z}$$

List them in ascending order (all in radians):

$$x_1 = \alpha$$

$$x_2 = \pi-\alpha$$

$$x_3 = 2\pi+\alpha$$

$$x_4 = 3\pi-\alpha$$

(and so on, two roots in every length-$$2\pi$$ interval).

The interval upper end is $$L=n\dfrac{\pi}{2}$$. To have exactly three solutions we need

$$2\pi+\alpha \;\le\;L\;\lt\;3\pi-\alpha\;-(1)$$

Compute the numerical bounds:

$$2\pi+\alpha \approx 6.2832+0.7297 = 7.0129$$

$$3\pi-\alpha \approx 9.4248-0.7297 = 8.6951$$

The factor $$\dfrac{\pi}{2}\approx1.5708$$. Divide the inequalities in $$(1)$$ by $$\dfrac{\pi}{2}$$:

$$\dfrac{7.0129}{1.5708}\;\le\;n\;\lt\;\dfrac{8.6951}{1.5708}$$

$$4.463\;\le\;n\;\lt\;5.533$$

With $$n\in\mathbb{N}$$, the only possible value is

$$n = 5$$

Step 3 : Analyse the quadratic with this $$n$$
Put $$n=5$$ in $$x^{2}+nx+(n-3)=0$$:

$$x^{2}+5x+2=0\;-(2)$$

Discriminant $$\Delta = 5^{2}-4\cdot1\cdot2 = 25-8 = 17$$

Roots of $$(2)$$ are

$$x=\dfrac{-5\pm\sqrt{17}}{2}$$

Numerically:

$$x_{1} = \dfrac{-5+\sqrt{17}}{2}\approx\dfrac{-5+4.123}{2}\approx-0.438$$

$$x_{2} = \dfrac{-5-\sqrt{17}}{2}\approx\dfrac{-5-4.123}{2}\approx-4.562$$

Both roots are negative, so they lie in the interval $$(-\infty,0)$$.

Answer
The roots of $$x^{2}+nx+(n-3)=0$$ belong to $$(-\infty,0)$$, that is Option B.

Rearrange and Square: $$4\cos\theta - 1 = 3\sin\theta$$

$$(4\cos\theta - 1)^2 = (3\sin\theta)^2$$

$$16\cos^2\theta - 8\cos\theta + 1 = 9(1 - \cos^2\theta)$$

$$16\cos^2\theta - 8\cos\theta + 1 = 9 - 9\cos^2\theta$$

$$25\cos^2\theta - 8\cos\theta - 8 = 0$$

Using Quadratic Formula

$$\cos\theta = \frac{8 \pm \sqrt{64 - 4(25)(-8)}}{2(25)} = \frac{8 \pm \sqrt{864}}{50} = \frac{8 \pm 12\sqrt{6}}{50} = \frac{4 \pm 6\sqrt{6}}{25}$$

Option C is $$\frac{4}{3\sqrt{6}-2}$$. Multiplying by the conjugate $$\frac{3\sqrt{6}+2}{3\sqrt{6}+2}$$:

$$\frac{4(3\sqrt{6}+2)}{54-4} = \frac{4(3\sqrt{6}+2)}{50} = \frac{6\sqrt{6}+4}{25}$$

This matches our positive root.

Given: $$3\sin(\alpha+\beta) = 2\sin(\alpha-\beta)$$.

Expanding: $$3(\sin\alpha\cos\beta + \cos\alpha\sin\beta) = 2(\sin\alpha\cos\beta - \cos\alpha\sin\beta)$$.

$$3\sin\alpha\cos\beta + 3\cos\alpha\sin\beta = 2\sin\alpha\cos\beta - 2\cos\alpha\sin\beta$$

$$\sin\alpha\cos\beta = -5\cos\alpha\sin\beta$$

$$\frac{\sin\alpha}{\cos\alpha} = \frac{-5\sin\beta}{\cos\beta}$$

$$\tan\alpha = -5\tan\beta$$

So $$k = -5$$.

The answer is Option (1): $$\boxed{-5}$$.

$$4 + 5 \tan \theta = \sec \theta$$.

Square both sides: $$(4 + 5 \tan \theta)^2 = \sec^2 \theta = 1 + \tan^2 \theta$$.

$$16 + 40 \tan \theta + 25 \tan^2 \theta = 1 + \tan^2 \theta$$.

$$24 \tan^2 \theta + 40 \tan \theta + 15 = 0$$.

Using the quadratic formula for $$\tan \alpha$$:

$$\tan \alpha = \frac{-40 \pm \sqrt{1600 - 4(24)(15)}}{2(24)} = \frac{-40 \pm \sqrt{1600 - 1440}}{48} = \frac{-40 \pm \sqrt{160}}{48} = \frac{-40 \pm 4\sqrt{10}}{48} = \frac{-10 \pm \sqrt{10}}{12}$$.

the value is $$\frac{\sqrt{10}-10}{12}$$

Given that $$\sin x = -\frac{3}{5}$$ and $$\pi < x < \frac{3\pi}{2}$$, which is the third quadrant. In this quadrant, sine and cosine are negative, and tangent is positive.

Using the Pythagorean identity, $$\sin^2 x + \cos^2 x = 1$$:

Substitute $$\sin x = -\frac{3}{5}$$:

$$\left(-\frac{3}{5}\right)^2 + \cos^2 x = 1$$

$$\frac{9}{25} + \cos^2 x = 1$$

$$\cos^2 x = 1 - \frac{9}{25} = \frac{16}{25}$$

Thus, $$\cos x = \pm \frac{4}{5}$$.

Since $$x$$ is in the third quadrant, cosine is negative, so $$\cos x = -\frac{4}{5}$$.

Now, find $$\tan x$$:

$$\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{3}{5}}{-\frac{4}{5}} = \frac{3}{4}$$.

Compute $$\tan^2 x - \cos x$$:

$$\tan^2 x = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$$

$$\cos x = -\frac{4}{5}$$

So, $$\tan^2 x - \cos x = \frac{9}{16} - \left(-\frac{4}{5}\right) = \frac{9}{16} + \frac{4}{5}$$

Add the fractions. The least common denominator of 16 and 5 is 80:

$$\frac{9}{16} = \frac{9 \times 5}{16 \times 5} = \frac{45}{80}$$

$$\frac{4}{5} = \frac{4 \times 16}{5 \times 16} = \frac{64}{80}$$

Thus, $$\frac{45}{80} + \frac{64}{80} = \frac{109}{80}$$

Now, compute $$80 \times (\tan^2 x - \cos x) = 80 \times \frac{109}{80} = 109$$.

The expression $$80(\tan^2 x - \cos x)$$ equals 109, which corresponds to option B.

Using the identity: $$\cos\theta \cos(60°-\theta)\cos(60°+\theta) = \frac{1}{4}\cos 3\theta$$.

So $$|\cos\theta \cos(60°-\theta)\cos(60°+\theta)| = \frac{1}{4}|\cos 3\theta| \leq \frac{1}{8}$$.

This gives $$|\cos 3\theta| \leq \frac{1}{2}$$, which is always true when $$|\cos 3\theta| \leq \frac{1}{2}$$.

Given the constraint $$|\cos 3\theta| \leq \frac{1}{2}$$, the maximum of $$\cos 3\theta$$ subject to this constraint is $$\frac{1}{2}$$.

$$\cos 3\theta = \frac{1}{2}$$ when $$3\theta = \frac{\pi}{3} + 2n\pi$$ or $$3\theta = -\frac{\pi}{3} + 2n\pi = \frac{5\pi}{3} + 2n\pi$$.

For $$\theta \in [0, 2\pi]$$, $$3\theta \in [0, 6\pi]$$:

$$3\theta = \frac{\pi}{3}$$; $$\theta = \frac{\pi}{9}$$

$$3\theta = \frac{5\pi}{3}$$; $$\theta = \frac{5\pi}{9}$$

$$3\theta = \frac{7\pi}{3}$$; $$\theta = \frac{7\pi}{9}$$

$$3\theta = \frac{11\pi}{3}$$; $$\theta = \frac{11\pi}{9}$$

$$3\theta = \frac{13\pi}{3}$$; $$\theta = \frac{13\pi}{9}$$

$$3\theta = \frac{17\pi}{3}$$; $$\theta = \frac{17\pi}{9}$$

Sum = $$\frac{\pi}{9}(1+5+7+11+13+17) = \frac{54\pi}{9} = 6\pi$$.

The correct answer is Option 3: $$6\pi$$.

Using the identity $$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$:

Let $$k = \sqrt{x^2+x+1}$$. Then $$\tan A = \frac{1}{\sqrt{x}k}$$ and $$\tan B = \frac{\sqrt{x}}{k}$$.

$$\tan(A+B) = \frac{\frac{1}{\sqrt{x}k} + \frac{\sqrt{x}}{k}}{1 - \frac{1}{k^2}} = \frac{\frac{1+x}{\sqrt{x}k}}{\frac{k^2-1}{k^2}} = \frac{(1+x)k}{\sqrt{x}(k^2-1)}$$.

Substitute $$k^2-1 = x^2+x$$:

$$\tan(A+B) = \frac{(1+x)\sqrt{x^2+x+1}}{\sqrt{x}(x^2+x)} = \frac{(1+x)\sqrt{x^2+x+1}}{x\sqrt{x}(1+x)} = \frac{\sqrt{x^2+x+1}}{x^{3/2}}$$.

This simplifies to $$\sqrt{\frac{x^2+x+1}{x^3}} = \sqrt{x^{-1} + x^{-2} + x^{-3}}$$.

Looking at $$\tan C$$, if we assume the expression in the image (which is slightly blurry but fits the pattern) represents the square root of the sum or is equivalent to this expansion, then $$\tan(A+B) = \tan C$$.

Since $$0 < A,B,C < \pi/2$$, then $$A+B = C$$.

We need to evaluate $$\frac{3\cos 36° + 5\sin 18°}{5\cos 36° - 3\sin 18°}$$.

Using the known values:

$$\cos 36° = \frac{\sqrt{5}+1}{4}, \quad \sin 18° = \frac{\sqrt{5}-1}{4}$$

$$3\cos 36° + 5\sin 18° = 3 \cdot \frac{\sqrt{5}+1}{4} + 5 \cdot \frac{\sqrt{5}-1}{4}$$

$$= \frac{3\sqrt{5}+3+5\sqrt{5}-5}{4} = \frac{8\sqrt{5}-2}{4} = \frac{4\sqrt{5}-1}{2}$$

$$5\cos 36° - 3\sin 18° = 5 \cdot \frac{\sqrt{5}+1}{4} - 3 \cdot \frac{\sqrt{5}-1}{4}$$

$$= \frac{5\sqrt{5}+5-3\sqrt{5}+3}{4} = \frac{2\sqrt{5}+8}{4} = \frac{\sqrt{5}+4}{2}$$

$$\frac{4\sqrt{5}-1}{2} \div \frac{\sqrt{5}+4}{2} = \frac{4\sqrt{5}-1}{\sqrt{5}+4}$$

Rationalizing by multiplying by $$\frac{\sqrt{5}-4}{\sqrt{5}-4}$$:

$$= \frac{(4\sqrt{5}-1)(\sqrt{5}-4)}{(\sqrt{5}+4)(\sqrt{5}-4)} = \frac{20-16\sqrt{5}-\sqrt{5}+4}{5-16} = \frac{24-17\sqrt{5}}{-11} = \frac{17\sqrt{5}-24}{11}$$

So the expression equals $$\frac{17\sqrt{5} - 24}{11}$$, where $$a = 17, b = 24, c = 11$$.

Checking: $$\gcd(17, 11) = 1$$. ✓

Therefore $$a + b + c = 17 + 24 + 11 = 52$$.

The correct answer is Option 2: 52.

We need to find the number of solutions of $$4\sin^2 x - 4\cos^3 x + 9 - 4\cos x = 0$$ in $$[-2\pi, 2\pi]$$.

$$4(1 - \cos^2 x) - 4\cos^3 x + 9 - 4\cos x = 0$$

$$4 - 4\cos^2 x - 4\cos^3 x + 9 - 4\cos x = 0$$

$$-4\cos^3 x - 4\cos^2 x - 4\cos x + 13 = 0$$

$$4\cos^3 x + 4\cos^2 x + 4\cos x - 13 = 0$$

$$4\cos^3 x + 4\cos^2 x + 4\cos x = 13 $$

We know that, $$\cos x \leq 1$$

$$\Rightarrow$$ $$4\cos^3 x \leq 4$$

$$\Rightarrow$$ $$4\cos^2 x \leq 4$$

$$\Rightarrow$$ $$4\cos x \leq 4$$

$$\therefore$$ $$4\cos^3 x + 4\cos^2 x + 4\cos x \leq 12$$

Thus, the maximum value of LHS can be $$12$$, and RHS = $$13$$

Thus, there is no solution for the equation between $$x \in [-2\pi, 2\pi]$$

Hence, option D is the correct choice.

We need to find the sum of solutions $$x \in \mathbb{R}$$ of the equation:

$$\frac{3\cos 2x + \cos^3 2x}{\cos^6 x - \sin^6 x} = x^3 - x^2 + 6.$$

First, simplify the denominator using the factorisation of difference of cubes:

$$\cos^6 x - \sin^6 x = (\cos^2 x)^3 - (\sin^2 x)^3 = (\cos^2 x - \sin^2 x)(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x) = \cos 2x \cdot [(\cos^2 x + \sin^2 x)^2 - \cos^2 x \sin^2 x] = \cos 2x \cdot \left[1 - \frac{\sin^2 2x}{4}\right] = \cos 2x \cdot \frac{4 - \sin^2 2x}{4} = \cos 2x \cdot \frac{4 - (1 - \cos^2 2x)}{4} = \cos 2x \cdot \frac{3 + \cos^2 2x}{4}.$$

Since the numerator is $$3\cos 2x + \cos^3 2x = \cos 2x(3 + \cos^2 2x),$$ the left-hand side becomes

$$\frac{\cos 2x(3 + \cos^2 2x)}{\cos 2x \cdot \frac{3 + \cos^2 2x}{4}} = \frac{1}{1/4} = 4,$$

provided $$\cos 2x \neq 0$$ and $$3 + \cos^2 2x \neq 0,$$ the latter holding since $$3 + \cos^2 2x \geq 3.$$

Therefore the equation reduces to

$$4 = x^3 - x^2 + 6,$$

which is equivalent to

$$x^3 - x^2 + 2 = 0.$$

Testing $$x = -1$$ gives $$(-1)^3 - (-1)^2 + 2 = -1 - 1 + 2 = 0,$$ so $$(x + 1)$$ is a factor. Polynomial division yields

$$x^3 - x^2 + 2 = (x + 1)(x^2 - 2x + 2).$$

The quadratic $$x^2 - 2x + 2 = 0$$ has discriminant $$4 - 8 = -4 < 0$$ and thus no real roots. The only real solution is $$x = -1.$$ Finally, since $$\cos 2(-1) = \cos(-2) = \cos 2 \approx -0.416 \neq 0,$$ the solution is valid.

The sum of all real solutions is $$-1$$.

Differentiate: $$f'(x) = 12\cos^2 x (-\sin x) + 6\sqrt{3} \cos x (-\sin x) = -6\sin x \cos x (2\cos x + \sqrt{3})$$.

Set $$f'(x) = 0$$:

o $$\sin x = 0 \implies x = \pi$$ (since $$0, 2\pi$$ are excluded).

o $$\cos x = 0 \implies x = \pi/2, 3\pi/2$$.

o $$\cos x = -\sqrt{3}/2 \implies x = 5\pi/6, 7\pi/6$$.

Test for Maxima (Sign change of $$f'(x)$$):

o At $$x = \pi/2$$: $$f'$$ changes from $$-$$ to $$+ \implies$$ Minima.

o At $$x = 5\pi/6$$: $$f'$$ changes from $$+$$ to $$- \implies$$ Maxima.

o At $$x = \pi$$: $$f'$$ changes from $$-$$ to $$+ \implies$$ Minima.

o At $$x = 7\pi/6$$: $$f'$$ changes from $$+$$ to $$- \implies$$ Maxima.

o At $$x = 3\pi/2$$: $$f'$$ changes from $$-$$ to $$+ \implies$$ Minima.

There are 2 points of local maxima ($$5\pi/6$$ and $$7\pi/6$$).

Correct Option: D

To begin, we find the range of the function $$f(x) = \frac{1}{7 - \sin 5x}$$.

The sine function satisfies $$-1 \leq \sin\theta \leq 1$$ for all real $$\theta$$, and since $$5x$$ spans all real values as $$x$$ ranges over $$\mathbb{R}$$, it follows that

$$-1 \leq \sin 5x \leq 1$$

The extreme values occur when $$\sin 5x = 1$$ (i.e.\ $$5x = \frac{\pi}{2} + 2k\pi$$) and $$\sin 5x = -1$$ (i.e.\ $$5x = -\frac{\pi}{2} + 2k\pi$$).

Next, we consider the denominator $$7 - \sin 5x$$. Multiplying the inequality $$-1 \leq \sin 5x \leq 1$$ by $$-1$$ reverses it, giving

$$-1 \leq -\sin 5x \leq 1$$

and adding 7 yields

$$6 \leq 7 - \sin 5x \leq 8$$

Since $$7 - \sin 5x$$ lies in $$[6,8]$$ and is always positive, we can take reciprocals. Because the function $$g(t) = \frac{1}{t}$$ is strictly decreasing for $$t > 0$$, reversing the inequality gives

$$\frac{1}{8} \leq \frac{1}{7 - \sin 5x} \leq \frac{1}{6}$$

We verify that these bounds are attained: when $$\sin 5x = -1$$, one finds $$f(x) = \frac{1}{7 - (-1)} = \frac{1}{8}$$, and when $$\sin 5x = 1$$, $$f(x) = \frac{1}{7 - 1} = \frac{1}{6}$$.

Therefore, the range of $$f(x)$$ is $$\left[\frac{1}{8}, \frac{1}{6}\right]$$. The correct answer is Option D: $$\left[\frac{1}{8}, \frac{1}{6}\right]$$.

$$f(x) = \frac{1}{2+\sin 3x+\cos 3x}$

Range of $$\sin 3x+\cos 3x = \sqrt{2}\sin(3x+\pi/4)$$ is $$[-\sqrt{2},\sqrt{2}]$$.

Denominator range: $$[2-\sqrt{2}, 2+\sqrt{2}]$$. So $$f$$ range: $$[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}]$$.

$$a = \frac{1}{2+\sqrt{2}} = \frac{2-\sqrt{2}}{2}$$, $$b = \frac{1}{2-\sqrt{2}} = \frac{2+\sqrt{2}}{2}$$.

AM: $$\alpha = \frac{a+b}{2} = \frac{(2-\sqrt{2}+2+\sqrt{2})/2}{2} = \frac{2}{2} = 1$$.

GM: $$\beta = \sqrt{ab} = \sqrt{\frac{(2-\sqrt{2})(2+\sqrt{2})}{4}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$$.

$$\alpha/\beta = 1/(1/\sqrt{2}) = \sqrt{2}$$

First, find $$[p, q]$$ for the equation $$\cos 2x + a\sin x = 2a - 7$$.

Using $$\cos 2x = 1 - 2\sin^2 x$$:

$$1 - 2\sin^2 x + a\sin x = 2a - 7$$

$$-2\sin^2 x + a\sin x + 8 - 2a = 0$$

$$2\sin^2 x - a\sin x + 2a - 8 = 0$$

Let $$t = \sin x \in [-1, 1]$$:

$$2t^2 - at + 2a - 8 = 0$$

$$a(2 - t) = 8 - 2t^2 = 2(4 - t^2) = 2(2-t)(2+t)$$

$$a = \frac{2(2-t)(2+t)}{2-t} = 2(2+t) = 4 + 2t$$ (for $$t \neq 2$$, which is always true since $$t \in [-1,1]$$).

For $$t \in [-1, 1]$$: $$a = 4 + 2t \in [2, 6]$$.

So $$p = 2$$, $$q = 6$$.

Now find $$r = \tan 9° - \tan 27° - \frac{1}{\cot 63°} + \tan 81°$$.

$$\frac{1}{\cot 63°} = \tan 63°$$.

$$r = \tan 9° - \tan 27° - \tan 63° + \tan 81°$$

$$= (\tan 9° + \tan 81°) - (\tan 27° + \tan 63°)$$

$$= (\tan 9° + \cot 9°) - (\tan 27° + \cot 27°)$$

$$= \frac{1}{\sin 9° \cos 9°} - \frac{1}{\sin 27° \cos 27°}$$

$$= \frac{2}{\sin 18°} - \frac{2}{\sin 54°}$$

$$= \frac{2}{\frac{\sqrt{5}-1}{4}} - \frac{2}{\frac{\sqrt{5}+1}{4}}$$

$$= \frac{8}{\sqrt{5}-1} - \frac{8}{\sqrt{5}+1}$$

$$= 8 \cdot \frac{(\sqrt{5}+1) - (\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} = 8 \cdot \frac{2}{4} = 4$$

$$pqr = 2 \times 6 \times 4 = 48$$

The answer is $$\boxed{48}$$.

We wish to find the number of solutions of $$\sin^2 x + (2+2x-x^2)\sin x - 3(x-1)^2 = 0$$ for $$-\pi \leq x \leq \pi$$.

Letting $$s = \sin x$$ and observing that $$2+2x-x^2 = 3-(x-1)^2$$ transforms the equation into $$s^2 + \bigl(3-(x-1)^2\bigr)s - 3(x-1)^2 = 0,$$ which can be rewritten as $$s^2 + 3s - (x-1)^2(s+3) = 0$$ and thus factors to $$(s+3)(s-(x-1)^2) = 0.$$

It follows that $$\sin x + 3 = 0$$, giving $$\sin x = -3$$, which is impossible since $$|\sin x|\leq1$$. Therefore we must have $$\sin x = (x-1)^2$$, requiring $$0 \leq (x-1)^2 \leq 1$$ and hence $$x \in [0,2].$$

On the interval $$[0,2]$$, comparing $$y = \sin x$$ and $$y = (x-1)^2$$ shows that at $$x = 0$$, $$\sin0 = 0$$ is below $$(0-1)^2 = 1$$; at $$x = 1$$, $$\sin1 \approx 0.841$$ is above $$(1-1)^2 = 0$$; and at $$x = 2$$, $$\sin2 \approx 0.909$$ is below $$(2-1)^2 = 1$$. Thus the two curves intersect twice, once in $$(0,1)$$ and once in $$(1,2)$$.

Therefore there are 2 solutions, and the correct answer is $$\boxed{2}$$.

In triangle $$ABC$$, $$BC = a = 7$$, $$AC = b = 8$$, $$AB = c = \alpha \in \mathbb{N}$$, and $$\cos A = 2/3$$.

Find $$\alpha$$ using the cosine rule.

$$a^2 = b^2 + c^2 - 2bc\cos A$$

$$49 = 64 + \alpha^2 - 2(8)(\alpha)(2/3) = 64 + \alpha^2 - \frac{32\alpha}{3}$$

$$\alpha^2 - \frac{32\alpha}{3} + 15 = 0$$

$$3\alpha^2 - 32\alpha + 45 = 0$$

$$\alpha = \frac{32 \pm \sqrt{1024 - 540}}{6} = \frac{32 \pm \sqrt{484}}{6} = \frac{32 \pm 22}{6}$$

$$\alpha = 9$$ or $$\alpha = 5/3$$ (not a natural number). So $$\alpha = 9$$.

Find $$\cos C$$ using the cosine rule.

$$c^2 = a^2 + b^2 - 2ab\cos C$$

$$81 = 49 + 64 - 2(7)(8)\cos C$$

$$81 = 113 - 112\cos C$$

$$\cos C = \frac{32}{112} = \frac{2}{7}$$

Compute $$\cos 3C$$.

$$\cos 3C = 4\cos^3 C - 3\cos C = 4\left(\frac{2}{7}\right)^3 - 3\left(\frac{2}{7}\right) = \frac{32}{343} - \frac{6}{7} = \frac{32 - 294}{343} = \frac{-262}{343}$$

Compute the expression.

$$49\cos 3C + 42 = 49 \cdot \frac{-262}{343} + 42 = \frac{-262}{7} + 42 = \frac{-262 + 294}{7} = \frac{32}{7}$$

So $$\frac{m}{n} = \frac{32}{7}$$ with $$\gcd(32, 7) = 1$$. Therefore $$m + n = 32 + 7 = 39$$.

The answer is 39.

Given,

$$\log_{\cos x}(\cot x) + 4\log_{\sin x}(\tan x) = 1$$

Using,

$$\cot x = \frac{\cos x}{\sin x}$$

and

$$\tan x = \frac{\sin x}{\cos x}$$

we get

$$\log_{\cos x}\left(\frac{\cos x}{\sin x}\right) + 4\log_{\sin x}\left(\frac{\sin x}{\cos x}\right) = 1$$

Using the logarithmic property

$$\log_a\left(\frac{m}{n}\right)=\log_a m-\log_a n$$

$$\log_{\cos x}\cos x - \log_{\cos x}\sin x + 4\left(\log_{\sin x}\sin x - \log_{\sin x}\cos x\right)=1$$

Since,

$$\log_{\cos x}\cos x = 1$$

and

$$\log_{\sin x}\sin x = 1$$

we obtain

$$1-\log_{\cos x}\sin x + 4\left(1-\log_{\sin x}\cos x\right)=1$$

Let

$$t=\log_{\cos x}\sin x$$

Using the reciprocal property of logarithms,

$$\log_{\sin x}\cos x = \frac{1}{t}$$

Substituting,

$$1-t+4\left(1-\frac{1}{t}\right)=1$$

$$1-t+4-\frac{4}{t}=1$$

$$5-t-\frac{4}{t}=1$$

$$4-t-\frac{4}{t}=0$$

Multiplying throughout by $$t$$,

$$4t-t^2-4=0$$

$$t^2-4t+4=0$$

$$(t-2)^2=0$$

$$t=2$$

Therefore,

$$\log_{\cos x}\sin x = 2$$

Converting to exponential form,

$$\sin x = \cos^2 x$$

Using,

$$\cos^2 x = 1-\sin^2 x$$

we get

$$\sin x = 1-\sin^2 x$$

$$\sin^2 x + \sin x -1 =0$$

Using the quadratic formula,

$$\sin x = \frac{-1\pm\sqrt{1+4}}{2}$$

$$\sin x = \frac{-1\pm\sqrt{5}}{2}$$

Therefore,

$$x=\sin^{-1}\left(\frac{-1+\sqrt5}{2}\right)$$

Comparing with

$$x=\sin^{-1}\left(\frac{\alpha+\sqrt\beta}{2}\right)$$

we get

$$\alpha=-1,\qquad \beta=5$$

Hence,

$$\alpha+\beta = -1+5 = 4$$

Therefore, the required answer is

$$\boxed{4}$$

We need to evaluate $$96 \cos\frac{\pi}{33} \cos\frac{2\pi}{33} \cos\frac{4\pi}{33} \cos\frac{8\pi}{33} \cos\frac{16\pi}{33}$$.

A well-known product formula for cosines with angles in geometric progression states that

$$\prod_{k=0}^{n-1}\cos(2^k\theta)=\frac{\sin(2^n\theta)}{2^n\sin\theta}$$ which follows by repeatedly applying the double-angle formula $$\sin(2\alpha)=2\sin\alpha\cos\alpha$$.

By setting $$\theta=\pi/33$$ and $$n=5$$ in this identity, it follows that

$$\cos\frac{\pi}{33}\cos\frac{2\pi}{33}\cos\frac{4\pi}{33}\cos\frac{8\pi}{33}\cos\frac{16\pi}{33} =\frac{\sin(2^5\cdot\pi/33)}{2^5\sin(\pi/33)} =\frac{\sin(32\pi/33)}{32\sin(\pi/33)}$$

Noting that $$\sin\frac{32\pi}{33}=\sin\bigl(\pi-\frac{\pi}{33}\bigr)=\sin\frac{\pi}{33}$$ by the identity $$\sin(\pi-x)=\sin x$$, the product simplifies to

$$\frac{\sin(\pi/33)}{32\sin(\pi/33)}=\frac{1}{32}$$

Finally, multiplying by 96 gives

$$96\times\frac{1}{32}=3$$ so the value of the given expression is 3.

Hence, the correct answer is Option 1: 3.

We need to find the range of $$\lambda$$ for which $$\cos^2 2x - 2\sin^4 x - 2\cos^2 x = \lambda$$ has solutions.

To begin,

Using $$\cos 2x = 1 - 2\sin^2 x$$:

$$\cos^2 2x = (1 - 2\sin^2 x)^2 = 1 - 4\sin^2 x + 4\sin^4 x$$

Substituting:

$$\lambda = 1 - 4\sin^2 x + 4\sin^4 x - 2\sin^4 x - 2\cos^2 x$$

Using $$\cos^2 x = 1 - \sin^2 x$$:

$$\lambda = 1 - 4\sin^2 x + 2\sin^4 x - 2(1 - \sin^2 x)$$

$$= 1 - 4\sin^2 x + 2\sin^4 x - 2 + 2\sin^2 x$$

$$= 2\sin^4 x - 2\sin^2 x - 1$$

Next,

Let $$t = \sin^2 x$$, where $$t \in [0, 1]$$.

$$\lambda = f(t) = 2t^2 - 2t - 1$$

From this,

$$f'(t) = 4t - 2 = 0 \implies t = \frac{1}{2}$$

Evaluating at critical point and endpoints:

$$f(0) = -1$$

$$f\left(\frac{1}{2}\right) = 2 \times \frac{1}{4} - 2 \times \frac{1}{2} - 1 = \frac{1}{2} - 1 - 1 = -\frac{3}{2}$$

$$f(1) = 2 - 2 - 1 = -1$$

Minimum value = $$-\frac{3}{2}$$, Maximum value = $$-1$$.

Therefore, $$\lambda \in \left[-\frac{3}{2}, -1\right]$$.

The correct answer is Option 4: $$\left[-\frac{3}{2}, -1\right]$$.

We need to find $$a + \frac{1}{a}$$ given that $$\tan 15° + \frac{1}{\tan 75°} + \frac{1}{\tan 105°} + \tan 195° = 2a$$.

To begin,

$$\frac{1}{\tan 75°} = \frac{1}{\cot 15°} = \tan 15°$$

$$\frac{1}{\tan 105°} = \frac{1}{\tan(180° - 75°)} = \frac{1}{-\tan 75°} = -\frac{1}{\tan 75°} = -\tan 15°$$

$$\tan 195° = \tan(180° + 15°) = \tan 15°$$

Next,

$$\tan 15° + \tan 15° - \tan 15° + \tan 15° = 2\tan 15°$$

So $$2a = 2\tan 15°$$, giving $$a = \tan 15°$$.

From this,

$$\tan 15° = 2 - \sqrt{3}$$

$$\frac{1}{\tan 15°} = \frac{1}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{(2-\sqrt{3})(2+\sqrt{3})} = \frac{2 + \sqrt{3}}{1} = 2 + \sqrt{3}$$

$$a + \frac{1}{a} = (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$$

The correct answer is Option 1: $$4$$.

We need to solve: $$3\cos^4\theta - 5\cos^2\theta - 2\sin^6\theta + 2 = 0$$

Let $$c = \cos^2\theta$$. Then $$\sin^2\theta = 1 - c$$.

$$\sin^6\theta = (1-c)^3 = 1 - 3c + 3c^2 - c^3$$

Substituting:

$$3c^2 - 5c - 2(1 - 3c + 3c^2 - c^3) + 2 = 0$$

$$3c^2 - 5c - 2 + 6c - 6c^2 + 2c^3 + 2 = 0$$

$$2c^3 - 3c^2 + c = 0$$

$$c(2c^2 - 3c + 1) = 0$$

$$c(2c - 1)(c - 1) = 0$$

So $$\cos^2\theta = 0$$, $$\cos^2\theta = \frac{1}{2}$$, or $$\cos^2\theta = 1$$.

Case 1: $$\cos^2\theta = 0$$ gives $$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$$ — 2 solutions

Case 2: $$\cos^2\theta = \frac{1}{2}$$ gives $$\cos\theta = \pm\frac{1}{\sqrt{2}}$$ — $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$ — 4 solutions

Case 3: $$\cos^2\theta = 1$$ gives $$\theta = 0, \pi, 2\pi$$ — 3 solutions

Total number of elements in set S = 2 + 4 + 3 = 9.

By setting $$t = 9^{\tan^2 x}$$ it follows that $$9^{1-\tan^2 x} = \frac{9}{t}$$. Hence the equation becomes $$\frac{9}{t} + t = 10 \implies t^2 - 10t + 9 = 0 \implies (t-1)(t-9) = 0$$ so that $$t=1$$ or $$t=9$$. If $$t=1$$ then $$\tan^2 x=0$$ giving $$x=0$$, while if $$t=9$$ then $$\tan^2 x=1$$ which leads to $$x=\pm\frac{\pi}{4}$$. Therefore the set of solutions is $$S = \left\{-\frac{\pi}{4}, 0, \frac{\pi}{4}\right\}$$.

The value of $$\beta$$ can be written as $$\beta = \tan^2\left(-\frac{\pi}{12}\right) + \tan^2(0) + \tan^2\left(\frac{\pi}{12}\right) = 2\tan^2\frac{\pi}{12}$$. Since $$\tan 15° = 2 - \sqrt{3}$$, it follows that $$\tan^2 15° = 7 - 4\sqrt{3}$$ and therefore $$\beta = 14 - 8\sqrt{3}$$.

Finally, evaluating the desired expression gives
$$\frac{1}{6}(\beta - 14)^2 = \frac{1}{6}(-8\sqrt{3})^2 = \frac{192}{6} = \textbf{32}$$

We use the identity: $$\frac{\sin 3\theta}{\sin\theta} = 3 - 4\sin^2\theta = 4\cos^2\theta - 1$$

Applying to each factor:

$$4\cos^2 9° - 1 = \frac{\sin 27°}{\sin 9°}$$

$$4\cos^2 27° - 1 = \frac{\sin 81°}{\sin 27°}$$

$$4\cos^2 81° - 1 = \frac{\sin 243°}{\sin 81°}$$

$$4\cos^2 243° - 1 = \frac{\sin 729°}{\sin 243°}$$

The product telescopes:

$$\frac{\sin 27°}{\sin 9°} \cdot \frac{\sin 81°}{\sin 27°} \cdot \frac{\sin 243°}{\sin 81°} \cdot \frac{\sin 729°}{\sin 243°} = \frac{\sin 729°}{\sin 9°}$$

Since $$729° = 2 \times 360° + 9°$$, we have $$\sin 729° = \sin 9°$$.

Therefore the product = $$\frac{\sin 9°}{\sin 9°} = 1$$

So $$36 \times 1 = 36$$, which is Option D.

We need to find when $$\cos 2A + \cos 2B + \cos 2C$$ is least for a triangle.

Using the identity: $$\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A \cos B \cos C$$

This is minimized when $$\cos A \cos B \cos C$$ is maximized, which occurs when $$A = B = C = 60°$$ (equilateral triangle). The minimum value is $$-1 - 4 \times \frac{1}{8} = -\frac{3}{2}$$.

For an equilateral triangle with inradius $$r = 3$$:

The relationship between inradius and side is: $$r = \frac{a}{2\sqrt{3}}$$ $$\Rightarrow a = 2\sqrt{3} \times 3 = 6\sqrt{3}$$

Checking each option:

Option 1: Perimeter = $$3 \times 6\sqrt{3} = 18\sqrt{3}$$. This is correct.

Option 2: For an equilateral triangle: $$\sin 2A + \sin 2B + \sin 2C = 3\sin 120° = \frac{3\sqrt{3}}{2}$$. Also $$\sin A + \sin B + \sin C = 3\sin 60° = \frac{3\sqrt{3}}{2}$$. These are equal. Correct.

Option 3: The distance from the incenter M to each vertex is $$MA = \frac{r}{\sin(A/2)} = \frac{3}{\sin 30°} = 6$$. The angle $$\angle AMB = 90° + \frac{C}{2} = 90° + 30° = 120°$$. Therefore: $$\vec{MA} \cdot \vec{MB} = |MA||MB|\cos(\angle AMB) = 6 \times 6 \times \cos 120° = 36 \times (-\frac{1}{2}) = -18$$. Correct.

Option 4: Area = $$\frac{\sqrt{3}}{4} \times (6\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 108 = 27\sqrt{3}$$. But the option states $$\frac{27\sqrt{3}}{2}$$, which is NOT correct.

Therefore, the statement that is NOT correct is option 4: area of $$\triangle ABC = \frac{27\sqrt{3}}{2}$$.

Trigonometry:

1. Distance ($$d$$): From $$\triangle\ $$ABQ, $$\tan 60^\circ = \frac{30}{d} \implies d = \frac{30}{\sqrt{3}} = 10\sqrt{3}$$.
2. Tower Height ($$h$$): From top $$A$$, $$\tan 15^\circ = \frac{30 - h}{d}$$.

   Using $$\tan 15^\circ = 2 - \sqrt{3}$$:

   $$30 - h = 10\sqrt{3}(2 - \sqrt{3}) = 20\sqrt{3} - 30 \implies \mathbf{h = 60 - 20\sqrt{3}}$$.

3. Area $$BCPQ$$: Since $$CB = PQ = h$$, it is a rectangle of sides $$d$$ and $$h$$

   $$\text{Area} = d \times h = 10\sqrt{3}(60 - 20\sqrt{3}) = 600\sqrt{3} - 600$$.

   Result: $$600(\sqrt{3} - 1) \text{ m}^2$$.

The height of the tower is $$h = 5$$ m. Person 1 stands due south with angle of elevation $$45°$$, and person 2 stands due west with angle of elevation $$30°$$.

$$ \tan 45° = \frac{h}{d_1} \Rightarrow 1 = \frac{5}{d_1} \Rightarrow d_1 = 5 \text{ m} $$

$$ \tan 30° = \frac{h}{d_2} \Rightarrow \frac{1}{\sqrt{3}} = \frac{5}{d_2} \Rightarrow d_2 = 5\sqrt{3} \text{ m} $$

Since south and west are perpendicular directions, the two persons and the base of the tower form a right triangle.

$$ D = \sqrt{d_1^2 + d_2^2} = \sqrt{25 + 75} = \sqrt{100} = 10 \text{ m} $$

The distance between the two persons is Option B: 10 m.

To solve this, we first simplify $$f(x)$$ using trigonometric identities.

1. Simplify $$f(x)$$

Recall that $$\sin x + \cos x = \sqrt{2}\sin(x + \frac{\pi}{4})$$ and $$\sin x - \cos x = -\sqrt{2}\cos(x + \frac{\pi}{4})$$.

$$f(x) = \frac{\sqrt{2}\sin(x + \frac{\pi}{4}) - \sqrt{2}}{-\sqrt{2}\cos(x + \frac{\pi}{4})} = \frac{1 - \sin(x + \frac{\pi}{4})}{\cos(x + \frac{\pi}{4})}$$

Using half-angle identities (let $$\theta = x + \frac{\pi}{4}$$), we get:

$$f(x) = \frac{1 - \cos(\frac{\pi}{2} - \theta)}{\sin(\frac{\pi}{2} - \theta)} = \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = \tan\left(\frac{\pi}{8} - \frac{x}{2}\right)$$

2. Find the Derivatives

Let $$u = \frac{\pi}{8} - \frac{x}{2}$$, so $$f(x) = \tan u$$.

• $$f'(x) = \sec^2 u \cdot (-\frac{1}{2})$$
• $$f''(x) = -\frac{1}{2} [2\sec u \cdot \sec u \tan u \cdot (-\frac{1}{2})] = \frac{1}{2} \sec^2 u \tan u$$

3. Evaluate at $$x = \frac{7\pi}{12}$$

Calculate $$u$$ first:

$$u = \frac{\pi}{8} - \frac{7\pi}{24} = \frac{3\pi - 7\pi}{24} = -\frac{4\pi}{24} = -\frac{\pi}{6}$$

Now find the values:

• $$f(-\frac{\pi}{6}) = \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}}$$
• $$f''(-\frac{\pi}{6}) = \frac{1}{2} \sec^2(-\frac{\pi}{6}) \tan(-\frac{\pi}{6}) = \frac{1}{2} \cdot (\frac{2}{\sqrt{3}})^2 \cdot (-\frac{1}{\sqrt{3}}) = \frac{1}{2} \cdot \frac{4}{3} \cdot (-\frac{1}{\sqrt{3}}) = -\frac{2}{3\sqrt{3}}$$

4. Final Calculation

$$f\left(\frac{7\pi}{12}\right) f''\left(\frac{7\pi}{12}\right) = \left(-\frac{1}{\sqrt{3}}\right) \left(-\frac{2}{3\sqrt{3}}\right) = \frac{2}{3 \cdot 3} = \mathbf{\frac{2}{9}}$$

We need to find $$mn$$ where $$m$$ and $$n$$ are the numbers of positive and negative values of $$\theta$$ in $$[-\pi, \pi]$$ satisfying $$\cos 2\theta \cos \frac{\theta}{2} = \cos 3\theta \cos \frac{9\theta}{2}$$.

Apply the product-to-sum formula.

Using $$2\cos A \cos B = \cos(A-B) + \cos(A+B)$$:

LHS: $$\cos 2\theta \cos \frac{\theta}{2} = \frac{1}{2}\left[\cos\frac{3\theta}{2} + \cos\frac{5\theta}{2}\right]$$

RHS: $$\cos 3\theta \cos \frac{9\theta}{2} = \frac{1}{2}\left[\cos\left(-\frac{3\theta}{2}\right) + \cos\frac{15\theta}{2}\right] = \frac{1}{2}\left[\cos\frac{3\theta}{2} + \cos\frac{15\theta}{2}\right]$$

Simplify.

Setting LHS = RHS:

This gives: $$\cos\frac{15\theta}{2} - \cos\frac{5\theta}{2} = 0$$

Using $$\cos C - \cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2}$$:

Solve each factor.

Case 1: $$\sin(5\theta) = 0 \implies 5\theta = n\pi \implies \theta = \frac{n\pi}{5}$$

In $$[-\pi, \pi]$$: $$n = -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5$$

Case 2: $$\sin\frac{5\theta}{2} = 0 \implies \frac{5\theta}{2} = n\pi \implies \theta = \frac{2n\pi}{5}$$

In $$[-\pi, \pi]$$: $$n = -2, -1, 0, 1, 2$$, giving $$\theta = -\frac{4\pi}{5}, -\frac{2\pi}{5}, 0, \frac{2\pi}{5}, \frac{4\pi}{5}$$

All of these are already included in Case 1.

Count positive and negative solutions.

The distinct solutions are $$\theta = \frac{n\pi}{5}$$ for $$n = -5, -4, ..., 4, 5$$.

Positive values ($$\theta > 0$$): $$\frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5}, \pi$$ → $$m = 5$$

Negative values ($$\theta < 0$$): $$-\frac{\pi}{5}, -\frac{2\pi}{5}, -\frac{3\pi}{5}, -\frac{4\pi}{5}, -\pi$$ → $$n = 5$$

Therefore: $$mn = 5 \times 5 = \boxed{25}$$.

We group the terms using complementary angles: $$\tan 81° = \cot 9°$$ and $$\tan 63° = \cot 27°$$, so the expression becomes $$(\tan 9° + \cot 9°) - (\tan 27° + \cot 27°)$$.

Using the identity $$\tan\theta + \cot\theta = \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{1}{\sin\theta\cos\theta} = \dfrac{2}{\sin 2\theta}$$, we get $$\dfrac{2}{\sin 18°} - \dfrac{2}{\sin 54°}$$.

Substituting $$\sin 18° = \dfrac{\sqrt{5}-1}{4}$$ and $$\sin 54° = \dfrac{\sqrt{5}+1}{4}$$, this becomes $$\dfrac{8}{\sqrt{5}-1} - \dfrac{8}{\sqrt{5}+1} = 8\cdot\dfrac{(\sqrt{5}+1)-(\sqrt{5}-1)}{(\sqrt{5})^2-1^2} = 8\cdot\dfrac{2}{4} = \boxed{4}$$.

From the figure:

• $$CD$$ is parallel to $$AB$$ (both are perpendicular to $$BE$$). Thus, $$CD = AB$$.
• In $$\triangle CAB$$: $$AC = AB \tan \theta_1$$.
• Area $$(\triangle CAB) = \frac{1}{2} \cdot AB \cdot AC = \frac{1}{2} (AB)^2 \tan \theta_1 = 2\sqrt{3} - 3$$.
• In $$\triangle CED$$: $$ED = CD \tan \theta_2 = AB \tan \theta_2$$.
• We are given $$\theta_1 + \theta_2 = \frac{\pi}{2}$$, which implies $$\theta_2 = \frac{\pi}{2} - \theta_1$$.
• Therefore, $$\tan \theta_2 = \cot \theta_1$$.

You are given that $$\sqrt{3}BE = 4AB$$, which means $$BE = \frac{4}{\sqrt{3}}AB$$.

From the geometry, $$BE = BD + DE$$.

Since $$BD = AC$$ (opposite sides of rectangle $$ACDB$$), we have:

$$BE = AB \tan \theta_1 + AB \cot \theta_1$$

$$\frac{4}{\sqrt{3}}AB = AB (\tan \theta_1 + \cot \theta_1)$$

$$\frac{4}{\sqrt{3}} = \frac{\sin \theta_1}{\cos \theta_1} + \frac{\cos \theta_1}{\sin \theta_1} = \frac{\sin^2 \theta_1 + \cos^2 \theta_1}{\sin \theta_1 \cos \theta_1} = \frac{1}{\sin \theta_1 \cos \theta_1}$$

$$\sin(2\theta_1) = 2 \sin \theta_1 \cos \theta_1 = 2 \left( \frac{\sqrt{3}}{4} \right) = \frac{\sqrt{3}}{2}$$

This gives two possible values for $$2\theta_1$$: $$60^\circ$$ or $$120^\circ$$.

• If $$2\theta_1 = 60^\circ \implies \theta_1 = 30^\circ$$ and $$\theta_2 = 60^\circ$$.
• If $$2\theta_1 = 120^\circ \implies \theta_1 = 60^\circ$$ and $$\theta_2 = 30^\circ$$.

The problem states $$\frac{\theta_2}{\theta_1}$$ is largest, so we choose $$\theta_1 = 30^\circ$$ and $$\theta_2 = 60^\circ$$.

Using the area of $$\triangle CAB$$:

$$\frac{1}{2} (AB)^2 \tan 30^\circ = 2\sqrt{3} - 3$$

$$\frac{1}{2} (AB)^2 \left( \frac{1}{\sqrt{3}} \right) = 2\sqrt{3} - 3$$

$$(AB)^2 = 2\sqrt{3}(2\sqrt{3} - 3) = 4(3) - 6\sqrt{3} = 12 - 6\sqrt{3}$$

Using the identity $$12 - 6\sqrt{3} = (3 - \sqrt{3})^2$$:

$$AB = 3 - \sqrt{3}$$

In $$\triangle CED$$ (a $$30^\circ-60^\circ-90^\circ$$ triangle with $$\theta_2 = 60^\circ$$):

• $$CD = AB = 3 - \sqrt{3}$$
• $$ED = CD \tan 60^\circ = (3 - \sqrt{3})\sqrt{3} = 3\sqrt{3} - 3$$
• $$CE = \frac{CD}{\cos 60^\circ} = 2(3 - \sqrt{3}) = 6 - 2\sqrt{3}$$

Perimeter $$= CD + ED + CE$$:

$$(3 - \sqrt{3}) + (3\sqrt{3} - 3) + (6 - 2\sqrt{3})$$

$$3 - 3 + 6 - \sqrt{3} + 3\sqrt{3} - 2\sqrt{3}$$

$$6 + 0\sqrt{3} = 6$$

Final Answer: 6

We need to find the value of $$16\sin(20°)\sin(40°)\sin(80°)$$.

A well-known product-to-sum identity states that $$\sin\theta \cdot \sin(60° - \theta) \cdot \sin(60° + \theta) = \frac{\sin 3\theta}{4}$$.

Substituting θ = 20° into this identity gives $$\sin(20°) \cdot \sin(40°) \cdot \sin(80°) = \sin(20°) \cdot \sin(60° - 20°) \cdot \sin(60° + 20°)$$.

It follows that $$= \frac{\sin(60°)}{4} = \frac{\sqrt{3}/2}{4} = \frac{\sqrt{3}}{8}$$.

Multiplying both sides by 16 yields $$16\sin(20°)\sin(40°)\sin(80°) = 16 \times \frac{\sqrt{3}}{8} = 2\sqrt{3}$$.

The answer is Option B: $$2\sqrt{3}$$.

We need to find $$n(S) + \displaystyle\sum_{\theta \in S} \sec\left(\frac{\pi}{4} + 2\theta\right)\csc\left(\frac{\pi}{4} + 2\theta\right)$$, where $$S = \{\theta \in [0, 2\pi] : 8^{2\sin^2\theta} + 8^{2\cos^2\theta} = 16\}$$.

Let $$t = 2\sin^2\theta$$, so $$2\cos^2\theta = 2 - t$$.

$$8^t + 8^{2-t} = 16$$

Let $$u = 8^t$$: $$u + \frac{64}{u} = 16$$, so $$u^2 - 16u + 64 = 0$$, giving $$(u - 8)^2 = 0$$.

Therefore $$u = 8$$, so $$8^t = 8^1$$, giving $$t = 1$$, i.e., $$\sin^2\theta = \frac{1}{2}$$.

$$\sin\theta = \pm\frac{1}{\sqrt{2}}$$, so $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$.

$$n(S) = 4$$

$$\sec\alpha \cdot \csc\alpha = \frac{1}{\sin\alpha \cos\alpha} = \frac{2}{\sin 2\alpha}$$, where $$\alpha = \frac{\pi}{4} + 2\theta$$.

For each $$\theta$$, we compute $$\sin(2\alpha) = \sin\left(\frac{\pi}{2} + 4\theta\right) = \cos(4\theta)$$:

$$\theta = \frac{\pi}{4}: \cos(\pi) = -1 \implies \text{value} = -2$$

$$\theta = \frac{3\pi}{4}: \cos(3\pi) = -1 \implies \text{value} = -2$$

$$\theta = \frac{5\pi}{4}: \cos(5\pi) = -1 \implies \text{value} = -2$$

$$\theta = \frac{7\pi}{4}: \cos(7\pi) = -1 \implies \text{value} = -2$$

Sum = $$-2 - 2 - 2 - 2 = -8$$

$$n(S) + \text{Sum} = 4 + (-8) = -4$$

Therefore, the correct answer is Option C: $$-4$$.

We need to find the number of solutions of $$ \cos x = |\sin x| $$ in the interval $$ -4\pi \le x \le 4\pi $$.

Since the RHS is $$ |\sin x| \geq 0 $$, we need $$ \cos x \geq 0 $$.

Squaring both sides (valid since both sides are non-negative when solutions exist):

$$\cos^2 x = \sin^2 x$$

$$\cos 2x = 0$$

$$2x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$

$$x = \frac{\pi}{4} + \frac{k\pi}{2}$$

The general solutions are $$ x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \ldots $$

We need $$ \cos x \geq 0 $$, which holds when $$ x \in [-\frac{\pi}{2} + 2k\pi, \frac{\pi}{2} + 2k\pi] $$.

At $$ x = \frac{\pi}{4} + \frac{k\pi}{2} $$:

• $$ k = 0 $$: $$ x = \frac{\pi}{4} $$, $$ \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} > 0 $$ ✓
• $$ k = 1 $$: $$ x = \frac{3\pi}{4} $$, $$ \cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2} < 0 $$ ✗
• $$ k = 2 $$: $$ x = \frac{5\pi}{4} $$, $$ \cos\frac{5\pi}{4} = -\frac{\sqrt{2}}{2} < 0 $$ ✗
• $$ k = 3 $$: $$ x = \frac{7\pi}{4} $$, $$ \cos\frac{7\pi}{4} = \frac{\sqrt{2}}{2} > 0 $$ ✓

So within each period of $$ 2\pi $$, we get exactly 2 valid solutions: $$ x = \frac{\pi}{4} + 2m\pi $$ and $$ x = -\frac{\pi}{4} + 2m\pi $$ for integer $$ m $$.

Solutions of the form $$ x = \frac{\pi}{4} + 2m\pi $$:

$$-4\pi \le \frac{\pi}{4} + 2m\pi \le 4\pi$$

$$-\frac{17}{8} \le m \le \frac{15}{8}$$

So $$ m = -2, -1, 0, 1 $$ — giving 4 solutions.

Solutions of the form $$ x = -\frac{\pi}{4} + 2m\pi $$:

$$-4\pi \le -\frac{\pi}{4} + 2m\pi \le 4\pi$$

$$-\frac{15}{8} \le m \le \frac{17}{8}$$

So $$ m = -1, 0, 1, 2 $$ — giving 4 solutions.

Total solutions = $$ 4 + 4 = 8 $$.

The number of solutions is 8, which corresponds to Option C.

We need to solve $$\cos\left(x + \frac{\pi}{3}\right)\cos\left(\frac{\pi}{3} - x\right) = \frac{1}{4}\cos^2 2x$$ in $$[-3\pi, 3\pi]$$. Simplify the left-hand side using the product-to-sum formula $$\cos A \cos B = \frac{1}{2}[\cos(A - B) + \cos(A + B)],$$ where $$A = x + \frac{\pi}{3}$$ and $$B = \frac{\pi}{3} - x$$, so that $$A - B = 2x,\quad A + B = \frac{2\pi}{3}$$ and hence $$\text{LHS} = \frac{1}{2}\left[\cos 2x + \cos\frac{2\pi}{3}\right] = \frac{1}{2}\left[\cos 2x - \frac{1}{2}\right].$$

The equation becomes $$\frac{1}{2}\left[\cos 2x - \frac{1}{2}\right] = \frac{1}{4}\cos^2 2x$$ and multiplying both sides by 4 gives $$2\cos 2x - 1 = \cos^2 2x.$$ Rearranging leads to $$\cos^2 2x - 2\cos 2x + 1 = 0,$$ i.e. $$(\cos 2x - 1)^2 = 0,$$ so $$\cos 2x = 1,$$ which in turn yields $$2x = 2n\pi$$ and hence $$x = n\pi$$ for integer $$n$$.

Within $$[-3\pi, 3\pi]$$ this gives $$x = -3\pi, -2\pi, -\pi, 0, \pi, 2\pi, 3\pi,$$ a total of 7 solutions. Therefore, the answer is Option D: 7.

We need to find the value of $$\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)$$.

Using the roots of unity approach:

The 7th roots of unity are the solutions of $$z^7 = 1$$, given by $$z = e^{i \cdot 2k\pi/7}$$ for $$k = 0, 1, 2, \ldots, 6$$.

The sum of all 7th roots of unity is zero:

$$\sum_{k=0}^{6} e^{i \cdot 2k\pi/7} = 0$$

Taking the real part of both sides:

$$\sum_{k=0}^{6} \cos\left(\frac{2k\pi}{7}\right) = 0$$

$$1 + \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right) + \cos\left(\frac{8\pi}{7}\right) + \cos\left(\frac{10\pi}{7}\right) + \cos\left(\frac{12\pi}{7}\right) = 0$$

Using the property $$\cos(2\pi - \theta) = \cos\theta$$:

$$\cos\left(\frac{12\pi}{7}\right) = \cos\left(2\pi - \frac{2\pi}{7}\right) = \cos\left(\frac{2\pi}{7}\right)$$

$$\cos\left(\frac{10\pi}{7}\right) = \cos\left(2\pi - \frac{4\pi}{7}\right) = \cos\left(\frac{4\pi}{7}\right)$$

$$\cos\left(\frac{8\pi}{7}\right) = \cos\left(2\pi - \frac{6\pi}{7}\right) = \cos\left(\frac{6\pi}{7}\right)$$

Substituting back:

$$1 + 2\left[\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)\right] = 0$$

Let $$S = \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)$$

$$1 + 2S = 0$$

$$S = -\frac{1}{2}$$

The correct answer is Option B: $$-\dfrac{1}{2}$$.

Since consecutive angles differ by $$\dfrac{\pi}{6}$$, we use the identity $$\sec A \cdot \sec B = \frac{1}{\cos A \cos B}$$ along with $$B - A = \dfrac{\pi}{6}$$ and $$\sin\!\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$$ to obtain $$\frac{\sin(B-A)}{\cos A \cos B} = \tan B - \tan A \implies \frac{1}{\cos A \cos B} = \frac{\tan B - \tan A}{\sin(B-A)} = 2(\tan B - \tan A).$$

Let $$\theta_m = \theta + (m-1)\dfrac{\pi}{6}$$. Then the given sum becomes $$\sum_{m=1}^{9} \sec\!\left(\theta + (m-1)\frac{\pi}{6}\right)\sec\!\left(\theta + \frac{m\pi}{6}\right) = 2\sum_{m=1}^{9}\left[\tan\!\left(\theta + \frac{m\pi}{6}\right) - \tan\!\left(\theta + \frac{(m-1)\pi}{6}\right)\right],$$ which telescopes.

After cancellation, only the first and last terms remain: $$2\left[\tan\!\left(\theta + \frac{9\pi}{6}\right) - \tan\theta\right] = 2\left[\tan\!\left(\theta + \frac{3\pi}{2}\right) - \tan\theta\right].$$

Using the periodicity of tangent ($$\tan$$ has period $$\pi$$) gives $$\tan\!\left(\theta + \frac{3\pi}{2}\right) = \tan\!\left(\theta + \frac{3\pi}{2} - \pi\right) = \tan\!\left(\theta + \frac{\pi}{2}\right) = -\cot\theta.$$ Hence the sum becomes $$2(-\cot\theta - \tan\theta) = -2\left(\frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta}\right) = -2 \cdot \frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta} = \frac{-2}{\sin\theta\cos\theta}.$$

Since $$\sin 2\theta = 2\sin\theta\cos\theta$$, this simplifies to $$\frac{-4}{\sin 2\theta}.$$

Setting this equal to $$\frac{-8}{\sqrt{3}}$$ yields $$\frac{-4}{\sin 2\theta} = \frac{-8}{\sqrt{3}} \implies \sin 2\theta = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}.$$ For $$\theta \in \left(0, \dfrac{\pi}{2}\right)$$, we have $$2\theta \in (0, \pi)$$, so $$2\theta = \frac{\pi}{3} \implies \theta = \frac{\pi}{6} \quad \text{or} \quad 2\theta = \frac{2\pi}{3} \implies \theta = \frac{\pi}{3},$$ giving $$S = \left\{\dfrac{\pi}{6},\ \dfrac{\pi}{3}\right\}$$.

Finally, $$\sum_{\theta \in S} \theta = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi + 2\pi}{6} = \boxed{\frac{\pi}{2}}.$$ The answer is Option C.

We need to evaluate $$2\sin\dfrac{\pi}{22}\sin\dfrac{3\pi}{22}\sin\dfrac{5\pi}{22}\sin\dfrac{7\pi}{22}\sin\dfrac{9\pi}{22}$$.

The angles here are of the form $$(2k-1)\pi/22$$ for $$k=1,2,3,4,5$$, and since $$22=2\times11$$ these can be viewed as $$(2k-1)\pi/(2\cdot11)\,$$.

Although there is a standard product identity for a positive integer $$n$$, namely

$$\prod_{k=1}^{n}\sin\dfrac{(2k-1)\pi}{4n} \;=\;\dfrac{\sqrt{2}}{2^n},$$

it does not directly apply here because our denominator is $$2\cdot11$$ rather than $$4n$$. Instead we use the identity valid for odd $$m$$,

$$\prod_{k=0}^{\frac{m-1}{2}-1}\sin\dfrac{(2k+1)\pi}{2m} \;=\;\dfrac{1}{2^{(m-1)/2}}.$$

Setting $$m=11$$ gives $$(m-1)/2=5$$, so

$$\sin\dfrac{\pi}{22}\,\sin\dfrac{3\pi}{22}\,\sin\dfrac{5\pi}{22}\,\sin\dfrac{7\pi}{22}\,\sin\dfrac{9\pi}{22} \;=\;\dfrac{1}{2^5}\;=\;\dfrac{1}{32}.\!$$

This evaluation can also be confirmed by examining the Chebyshev polynomial $$U_{10}(x)$$, since $$U_{10}(\cos\theta)=\dfrac{\sin(11\theta)}{\sin\theta}$$ has roots at $$x=\cos\frac{k\pi}{11}$$ for $$k=1,2,\dots,10$$ and leading coefficient $$2^{10}$$. The factorization of $$U_{10}$$ yields the same product of odd‐indexed sines as $$1/32$$.

Finally, multiplying by 2 gives $$2\times\dfrac{1}{32}=\dfrac{1}{16}\,$$, so the correct answer is Option A: $$\dfrac{1}{16}$$.

We need to find which equation has $$\alpha = \sin 36°$$ as a root.

Use the key identity $$5 \times 36° = 180°$$

Since $$5 \times 36° = 180°$$, we have $$\sin(5 \times 36°) = \sin 180° = 0$$.

Let $$\theta = 36°$$. Then $$\sin 5\theta = 0$$.

Expand $$\sin 5\theta$$ using the multiple angle formula

The formula for $$\sin 5\theta$$ in terms of $$\sin\theta$$ is:

$$\sin 5\theta = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta$$

To derive this, we use:

$$\sin 5\theta = \sin(3\theta + 2\theta) = \sin 3\theta \cos 2\theta + \cos 3\theta \sin 2\theta$$

Where:

$$\sin 3\theta = 3\sin\theta - 4\sin^3\theta$$

$$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$$

$$\sin 2\theta = 2\sin\theta\cos\theta$$

$$\cos 2\theta = 1 - 2\sin^2\theta$$

Substituting:

$$\sin 5\theta = (3\sin\theta - 4\sin^3\theta)(1 - 2\sin^2\theta) + (4\cos^3\theta - 3\cos\theta)(2\sin\theta\cos\theta)$$

First part: $$(3\sin\theta - 4\sin^3\theta)(1 - 2\sin^2\theta)$$

$$= 3\sin\theta - 6\sin^3\theta - 4\sin^3\theta + 8\sin^5\theta$$

$$= 3\sin\theta - 10\sin^3\theta + 8\sin^5\theta$$

Second part: $$(4\cos^3\theta - 3\cos\theta)(2\sin\theta\cos\theta)$$

$$= 2\sin\theta\cos\theta(4\cos^3\theta - 3\cos\theta)$$

$$= 2\sin\theta(4\cos^4\theta - 3\cos^2\theta)$$

Using $$\cos^2\theta = 1 - \sin^2\theta$$:

$$\cos^4\theta = (1-\sin^2\theta)^2 = 1 - 2\sin^2\theta + \sin^4\theta$$

$$= 2\sin\theta[4(1-2\sin^2\theta+\sin^4\theta) - 3(1-\sin^2\theta)]$$

$$= 2\sin\theta[4 - 8\sin^2\theta + 4\sin^4\theta - 3 + 3\sin^2\theta]$$

$$= 2\sin\theta[1 - 5\sin^2\theta + 4\sin^4\theta]$$

$$= 2\sin\theta - 10\sin^3\theta + 8\sin^5\theta$$

Adding both parts:

$$\sin 5\theta = (3\sin\theta - 10\sin^3\theta + 8\sin^5\theta) + (2\sin\theta - 10\sin^3\theta + 8\sin^5\theta)$$

$$= 5\sin\theta - 20\sin^3\theta + 16\sin^5\theta$$

$$= \sin\theta(16\sin^4\theta - 20\sin^2\theta + 5)$$

Set $$\sin 5\theta = 0$$ and simplify

$$\sin\theta(16\sin^4\theta - 20\sin^2\theta + 5) = 0$$

Since $$\theta = 36°$$, $$\sin\theta \neq 0$$. So we divide by $$\sin\theta$$:

$$16\sin^4\theta - 20\sin^2\theta + 5 = 0$$

Write in terms of $$x = \sin 36°$$

$$16x^4 - 20x^2 + 5 = 0$$

Verify

We know $$\sin 36° = \frac{\sqrt{10-2\sqrt{5}}}{4}$$, so $$\sin^2 36° = \frac{10-2\sqrt{5}}{16}$$.

$$16x^4 = 16 \times \frac{(10-2\sqrt{5})^2}{256} = \frac{100 - 40\sqrt{5} + 20}{16} = \frac{120-40\sqrt{5}}{16}$$

$$20x^2 = \frac{200-40\sqrt{5}}{16}$$

$$16x^4 - 20x^2 + 5 = \frac{120-40\sqrt{5} - 200+40\sqrt{5}}{16} + 5 = \frac{-80}{16} + 5 = -5 + 5 = 0$$ ✓

Therefore, $$\sin 36°$$ is a root of $$16x^4 - 20x^2 + 5 = 0$$.

The correct answer is Option A: $$16x^4 - 20x^2 + 5 = 0$$.

Given: $$\cot \alpha = 1$$ with $$\pi < \alpha < \frac{3\pi}{2}$$, and $$\sec \beta = -\frac{5}{3}$$ with $$\frac{\pi}{2} < \beta < \pi$$.

Find $$\tan \alpha$$ and $$\tan \beta$$:

Since $$\cot \alpha = 1$$, we get $$\tan \alpha = 1$$.

(In the third quadrant, both sine and cosine are negative, so tangent is positive. This is consistent.)

Since $$\sec \beta = -\frac{5}{3}$$, we get $$\cos \beta = -\frac{3}{5}$$.

In the second quadrant, $$\sin \beta > 0$$, so $$\sin \beta = \frac{4}{5}$$.

$$\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/5}{-3/5} = -\frac{4}{3}$$

Compute $$\tan(\alpha + \beta)$$:

$$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta} = \frac{1 + \left(-\frac{4}{3}\right)}{1 - (1)\left(-\frac{4}{3}\right)} = \frac{-\frac{1}{3}}{1 + \frac{4}{3}} = \frac{-\frac{1}{3}}{\frac{7}{3}} = -\frac{1}{7}$$

Determine the quadrant of $$(\alpha + \beta)$$:

Since $$\alpha = \frac{5\pi}{4}$$ (the angle in the third quadrant with $$\tan \alpha = 1$$) and $$\beta = \pi - \arctan\!\left(\frac{4}{3}\right)$$ (in the second quadrant), we get:

$$\alpha + \beta = \frac{5\pi}{4} + \pi - \arctan\!\left(\frac{4}{3}\right) \approx 3.927 + 2.214 \approx 6.14$$

Since $$\frac{3\pi}{2} \approx 4.71$$ and $$2\pi \approx 6.28$$, we have $$\frac{3\pi}{2} < \alpha + \beta < 2\pi$$.

This places $$\alpha + \beta$$ in the fourth quadrant, which is consistent with $$\tan(\alpha + \beta) = -\frac{1}{7} < 0$$.

The correct answer is Option A: $$-\dfrac{1}{7}$$ and IV$$^{\text{th}}$$ quadrant.

We need to solve: $$\sin\theta\tan\theta + \tan\theta = \sin 2\theta$$ for $$\theta \in [-\pi, \pi] \setminus \{\pm\frac{\pi}{2}\}$$.

Rewriting the equation:

$$ \tan\theta(\sin\theta + 1) = 2\sin\theta\cos\theta $$

$$ \frac{\sin\theta}{\cos\theta}(\sin\theta + 1) = 2\sin\theta\cos\theta $$

Case 1: $$\sin\theta = 0$$

LHS = 0, RHS = 0. This gives $$\theta = -\pi, 0, \pi$$.

Note: $$\theta = -\pi$$ and $$\theta = \pi$$ represent the same point. We include both if both are in the domain. Since the domain is $$[-\pi, \pi]$$, both $$-\pi$$ and $$\pi$$ are included.

So $$\theta \in \{-\pi, 0, \pi\}$$.

Case 2: $$\sin\theta \neq 0$$. Divide both sides by $$\sin\theta$$:

$$ \frac{\sin\theta + 1}{\cos\theta} = 2\cos\theta $$

$$ \sin\theta + 1 = 2\cos^2\theta = 2(1 - \sin^2\theta) = 2 - 2\sin^2\theta $$

$$ 2\sin^2\theta + \sin\theta - 1 = 0 $$

$$ (2\sin\theta - 1)(\sin\theta + 1) = 0 $$

$$ \sin\theta = \frac{1}{2} \quad \text{or} \quad \sin\theta = -1 $$

$$\sin\theta = -1$$ gives $$\theta = -\frac{\pi}{2}$$, which is excluded from the domain.

$$\sin\theta = \frac{1}{2}$$ gives $$\theta = \frac{\pi}{6}$$ or $$\theta = \frac{5\pi}{6}$$.

We must verify $$\cos\theta \neq 0$$ for these values (already satisfied).

So $$S = \left\{-\pi, 0, \pi, \frac{\pi}{6}, \frac{5\pi}{6}\right\}$$

$$n(S) = 5$$

Now compute $$T = \sum_{\theta \in S} \cos 2\theta$$:

$$\cos 2(-\pi) = \cos(-2\pi) = 1$$

$$\cos 2(0) = \cos 0 = 1$$

$$\cos 2(\pi) = \cos 2\pi = 1$$

$$\cos 2\left(\frac{\pi}{6}\right) = \cos\frac{\pi}{3} = \frac{1}{2}$$

$$\cos 2\left(\frac{5\pi}{6}\right) = \cos\frac{5\pi}{3} = \frac{1}{2}$$

$$ T = 1 + 1 + 1 + \frac{1}{2} + \frac{1}{2} = 4 $$

$$ T + n(S) = 4 + 5 = 9 $$

The answer is Option D: 9.

We need to find the number of elements in $$S = \left\{x \in \mathbb{R} : 2\cos\left(\dfrac{x^2+x}{6}\right) = 4^x + 4^{-x}\right\}$$.

We observe that the right-hand side can be written as $$4^x + 4^{-x} = 2^{2x} + 2^{-2x}$$. By the AM-GM inequality, $$2^{2x} + 2^{-2x} \geq 2\sqrt{2^{2x} \cdot 2^{-2x}} = 2$$. Equality holds if and only if $$2^{2x} = 2^{-2x}$$, i.e., $$x = 0$$.

Now for the left-hand side, we know that $$\cos\theta \leq 1$$ for all real $$\theta$$, so $$2\cos\left(\dfrac{x^2+x}{6}\right) \leq 2$$.

Since the LHS $$\leq 2$$ and the RHS $$\geq 2$$, equality is possible only when both sides equal 2 simultaneously. The RHS equals 2 only when $$x = 0$$. At $$x = 0$$, the LHS becomes $$2\cos\left(\dfrac{0+0}{6}\right) = 2\cos(0) = 2$$.

Since both sides equal 2 at $$x = 0$$ and no other value of $$x$$ can satisfy the equation, there is exactly 1 element in the set $$S$$.

Hence, the correct answer is Option A: $$1$$.

We have a horizontal triangular park OAB with $$AB = 16$$ and a vertical lamp post OP at point O. C is the midpoint of AB, so $$AC = CB = 8$$. We are given $$\angle PAO = \angle PBO = 15°$$ and $$\angle PCO = 45°$$.

Since OP is vertical, the angles of elevation from A, B, and C to P are the given angles. Let $$OP = h$$. Then:

Since $$OA = OB$$, triangle OAB is isosceles with O on the perpendicular bisector of AB. Therefore O lies on the line through C perpendicular to AB... actually, O lies on the perpendicular bisector of AB, which passes through C (the midpoint of AB). So O, C are on the same line perpendicular to AB, meaning $$OC$$ is the distance from O to the midpoint C along this perpendicular bisector.

In triangle OAC (right-angled at C, since OC is perpendicular to AB):

Now $$\tan 15° = 2 - \sqrt{3}$$, so $$\tan^2 15° = (2-\sqrt{3})^2 = 7 - 4\sqrt{3}$$.

And $$1 - \tan^2 15° = 1 - 7 + 4\sqrt{3} = -6 + 4\sqrt{3}$$.

So:

Rationalizing by multiplying by $$\frac{2\sqrt{3}+3}{2\sqrt{3}+3}$$:

Expanding the numerator factor: $$(7 - 4\sqrt{3})(2\sqrt{3}+3) = 14\sqrt{3} + 21 - 24 - 12\sqrt{3} = 2\sqrt{3} - 3$$.

So:

Hence, the correct answer is Option 2.

Since $$\sin 12° = \sin(30° - 18°) = \sin 30° \cos 18° - \cos 30° \sin 18°$$, it follows that $$\sin 12° = \frac{1}{2}\cos 18° - \frac{\sqrt{3}}{2}\sin 18°$$.

Substituting this into $$2\sin 12° - \sin 72°$$ and using $$\sin 72° = \cos 18°$$ gives $$2\sin 12° - \sin 72° = 2\Bigl(\frac{1}{2}\cos 18° - \frac{\sqrt{3}}{2}\sin 18°\Bigr) - \cos 18° = \cos 18° - \sqrt{3}\sin 18° - \cos 18° = -\sqrt{3}\sin 18°$$.

Since $$\sin 18° = \frac{\sqrt{5} - 1}{4}$$, this yields $$2\sin 12° - \sin 72° = -\sqrt{3} \cdot \frac{\sqrt{5} - 1}{4} = \frac{\sqrt{3}(1 - \sqrt{5})}{4}$$. Therefore, the answer is $$\frac{\sqrt{3}(1 - \sqrt{5})}{4}$$ (Option D).

Let $$AB$$ and $$PQ$$ be two vertical poles standing 160 m apart, with heights $$AB = h$$ and $$PQ = 2h$$. Point $$C$$ is the midpoint of the segment joining the feet $$B$$ and $$Q$$ of the poles, so that $$BC = CQ = 80$$ m.

Since the angle of elevation from $$C$$ to the top $$P$$ of the taller pole is $$\frac{\pi}{8}$$, and the vertical height of that pole is $$PQ = 2h$$ over a horizontal distance $$CQ = 80$$ m, we have

$$\tan\left(\frac{\pi}{8}\right) = \frac{PQ}{CQ} = \frac{2h}{80},$$

which gives

$$h = 40\tan\left(\frac{\pi}{8}\right)\quad\cdots(1)$$

To evaluate $$\tan\left(\frac{\pi}{8}\right)$$, set $$t = \tan\left(\frac{\pi}{8}\right)$$. Noting that $$\frac{\pi}{8} = \tfrac12\cdot\tfrac{\pi}{4}$$ and applying the double‐angle formula,

$$\tan\left(\frac{\pi}{4}\right) = \frac{2t}{1 - t^2} = 1,$$

so

$$1 - t^2 = 2t,\quad t^2 + 2t - 1 = 0.$$

By the quadratic formula one finds

$$t = \frac{-2 \pm \sqrt{4 + 4}}{2} = -1 \pm \sqrt{2},$$

and since $$t>0$$ in the first quadrant,

$$\tan\left(\frac{\pi}{8}\right) = \sqrt{2} - 1.$$

Substituting this into (1) yields

$$h = 40(\sqrt{2} - 1).$$

Now, let $$\theta$$ be the angle of elevation from $$C$$ to the top $$A$$ of the shorter pole. Since $$AB = h$$ and $$BC = 80$$ m,

$$\tan\theta = \frac{AB}{BC} = \frac{h}{80} = \frac{40(\sqrt{2} - 1)}{80} = \frac{\sqrt{2} - 1}{2}.$$

Therefore,

$$\tan^2\theta = \left(\frac{\sqrt{2} - 1}{2}\right)^2 = \frac{(\sqrt{2} - 1)^2}{4},$$

and since $$(\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2},$$

we get

$$\tan^2\theta = \frac{3 - 2\sqrt{2}}{4}.$$

Hence the correct answer is Option C: $$\dfrac{3 - 2\sqrt{2}}{4}$$.

A tower $$ PQ $$ stands on the ground with base $$ Q $$. Point $$ R $$ divides the tower such that $$ QR = 15 $$ m. From point $$ A $$ on the ground, the angle of elevation of $$ R $$ is $$ 60° $$ and the part $$ PR $$ subtends an angle of $$ 15° $$ at $$ A $$.

Let the height of the tower be $$ h = PQ $$, so $$ PR = h - 15 $$.

Let $$ d = AQ $$ (horizontal distance from A to the base).

The angle of elevation of $$ R $$ from $$ A $$ is $$ 60° $$, so:

$$\tan 60° = \frac{QR}{d} = \frac{15}{d}$$

$$d = \frac{15}{\tan 60°} = \frac{15}{\sqrt{3}} = 5\sqrt{3}$$

The angle of elevation of $$ P $$ from $$ A $$ is $$ 60° + 15° = 75° $$.

$$\tan 75° = \frac{h}{d} = \frac{h}{5\sqrt{3}}$$

$$\tan 75° = \tan(45° + 30°) = \frac{\tan 45° + \tan 30°}{1 - \tan 45° \cdot \tan 30°} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$

Rationalizing:

$$\tan 75° = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$

$$h = 5\sqrt{3} \cdot \tan 75° = 5\sqrt{3}(2 + \sqrt{3}) = 10\sqrt{3} + 5 \cdot 3 = 10\sqrt{3} + 15$$

$$h = 5(2\sqrt{3} + 3) \text{ m}$$

The height of the tower is $$ 5(2\sqrt{3} + 3) $$ m, which corresponds to Option A.

We need to find $$\frac{R}{r}$$ for a triangle whose sides satisfy $$\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9}$$.

Let the common value be $$k$$, so that $$\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9} = k$$. It follows that
$$a + b = 7k,\quad b + c = 8k,\quad c + a = 9k$$ and adding these gives $$2(a + b + c) = 24k\implies a + b + c = 12k$$. Hence $$c = 12k - 7k = 5k$$, $$a = 12k - 8k = 4k$$, and $$b = 12k - 9k = 3k$$.

Since the sides are $$3k, 4k, 5k$$ and $$3^2 + 4^2 = 5^2$$, the triangle is right-angled with hypotenuse $$c = 5k$$.

In a right triangle, the circumradius is half the hypotenuse, so $$R = \frac{c}{2} = \frac{5k}{2}$$.

Moreover, the semi-perimeter is $$s = \frac{a + b + c}{2} = \frac{12k}{2} = 6k$$ and the area is $$\Delta = \frac{1}{2}\times a \times b = \frac{1}{2}\times 3k \times 4k = 6k^2$$ (using the two legs). Therefore, $$r = \frac{\Delta}{s} = \frac{6k^2}{6k} = k$$.

It follows that $$\frac{R}{r} = \frac{5k/2}{k} = \frac{5}{2}$$.

The correct answer is Option C: $$\dfrac{5}{2}$$.

Let the base of the tower be at point $$T$$ and its height be $$h$$. Point $$A$$ is due north of the tower, and point $$B$$ is 9 units due west of $$A$$.

We place $$T$$ at the origin. Since $$A$$ is due north of $$T$$, let the distance $$TA = d$$. Since $$B$$ is 9 units due west of $$A$$, the distance $$TB = \sqrt{d^2 + 81}$$.

We are given that the distance from $$B$$ to the tower (i.e., $$TB$$) is 15 units. So $$\sqrt{d^2 + 81} = 15$$, giving $$d^2 + 81 = 225$$, hence $$d^2 = 144$$ and $$d = 12$$.

Now, the angle of elevation from $$B$$ is $$\beta = \cos^{-1}\!\left(\dfrac{3}{\sqrt{13}}\right)$$. We have $$\cos\beta = \dfrac{3}{\sqrt{13}}$$, so $$\sin\beta = \dfrac{2}{\sqrt{13}}$$ and $$\tan\beta = \dfrac{2}{3}$$.

Since $$\tan\beta = \dfrac{h}{TB} = \dfrac{h}{15}$$, we get $$h = 15 \times \dfrac{2}{3} = 10$$.

The angle of elevation from $$A$$ is $$\alpha$$, with $$\tan\alpha = \dfrac{h}{d} = \dfrac{10}{12} = \dfrac{5}{6}$$.

Hence $$\cot\alpha = \dfrac{1}{\tan\alpha} = \dfrac{6}{5}$$.

Hence, the correct answer is Option A.

Let Q be at the origin with the horizontal ground along the x-axis. The tower PQ has height 10, so P = (0, 10) and Q = (0, 0).

Point A is on the ground with the angle of elevation of P being 45°:

So A = (10, 0).

Point R lies on segment AQ, so R = (x, 0) for some 0 ≤ x ≤ 10, and B is vertically above R, so B = (x, h) for some height h. Given ∠BAQ = 30° and AR = 10 − x,

From B = (x, h), the angle of elevation of P = (0, 10) is 60°:

Hence

The distance AB is

PQ and BR are vertical (parallel), with PQ = 10 and BR = 5($$\sqrt{3}$$-1). The horizontal distance between them is QR = x = 5($$\sqrt{3}$$-1). Therefore, the area of the trapezoid is

Therefore,

The answer is Option A.

Let the pole be at point A (base) with height 20 m (top at point P) and the tower be at point B (base) with height $$h$$ (top at point T). From the base of the pole (A), the angle of elevation to the top of the tower (T) is 60°. Let the horizontal distance between the pole and tower be $$d$$. Then $$\tan 60° = \frac{h}{d} \Rightarrow h = d\sqrt{3}$$.

The pole subtends an angle of 30° at the top of the tower T. From T, looking at the pole which extends from A (ground level) to P (height 20 m), the angle subtended is $$\angle ATP = 30°$$. Let $$\alpha$$ be the angle of depression from T to A and $$\beta$$ be the angle of depression from T to P, so that $$\tan \alpha = \frac{h}{d}$$ and $$\tan \beta = \frac{h - 20}{d}$$. The angle subtended by the pole at T is $$\alpha - \beta = 30°$$.

By the tangent subtraction formula, $$\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} = \tan 30° = \frac{1}{\sqrt{3}}$$, which gives $$\frac{\frac{h}{d} - \frac{h-20}{d}}{1 + \frac{h}{d}\cdot\frac{h-20}{d}} = \frac{1}{\sqrt{3}}$$, then $$\frac{\frac{20}{d}}{1 + \frac{h(h-20)}{d^2}} = \frac{1}{\sqrt{3}}$$ and hence $$\frac{20d}{d^2 + h(h-20)} = \frac{1}{\sqrt{3}}$$.

Substituting $$h = d\sqrt{3}$$ into the denominator yields $$d^2 + h(h-20) = d^2 + d\sqrt{3}(d\sqrt{3} - 20) = d^2 + 3d^2 - 20d\sqrt{3} = 4d^2 - 20d\sqrt{3}$$, so that $$\frac{20d}{4d^2 - 20d\sqrt{3}} = \frac{1}{\sqrt{3}}$$. Simplifying gives $$\frac{20}{4d - 20\sqrt{3}} = \frac{1}{\sqrt{3}}$$, hence $$20\sqrt{3} = 4d - 20\sqrt{3}$$, $$4d = 40\sqrt{3}$$ and $$d = 10\sqrt{3}$$.

Finally, $$h = d\sqrt{3} = 10\sqrt{3} \times \sqrt{3} = 30$$. Therefore, the height of the tower is 30 meters.

We need to find the sum of common solutions of $$2\sin^2\theta - \cos 2\theta = 0$$ and $$2\cos^2\theta + 3\sin\theta = 0$$ in $$[0, 2\pi]$$.

$$2\sin^2\theta - \cos 2\theta = 0$$

Using $$\cos 2\theta = 1 - 2\sin^2\theta$$:

$$2\sin^2\theta - (1 - 2\sin^2\theta) = 0$$

$$4\sin^2\theta - 1 = 0$$

$$\sin^2\theta = \dfrac{1}{4}$$

$$\sin\theta = \pm\dfrac{1}{2}$$

Solutions in $$[0, 2\pi]$$: $$\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$$

$$2\cos^2\theta + 3\sin\theta = 0$$

Using $$\cos^2\theta = 1 - \sin^2\theta$$:

$$2(1 - \sin^2\theta) + 3\sin\theta = 0$$

$$2 - 2\sin^2\theta + 3\sin\theta = 0$$

$$2\sin^2\theta - 3\sin\theta - 2 = 0$$

$$(2\sin\theta + 1)(\sin\theta - 2) = 0$$

Since $$\sin\theta \neq 2$$: $$\sin\theta = -\dfrac{1}{2}$$

Solutions in $$[0, 2\pi]$$: $$\theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$$

Common solutions: $$\theta = \dfrac{7\pi}{6}$$ and $$\theta = \dfrac{11\pi}{6}$$

$$\dfrac{7\pi}{6} + \dfrac{11\pi}{6} = \dfrac{18\pi}{6} = 3\pi$$

So $$k\pi = 3\pi$$, giving $$k = 3$$.

The correct answer is $$\boxed{3}$$.

We first solve $$7\cos^2\theta - 3\sin^2\theta - 2\cos^2(2\theta) = 2$$ for $$\theta \in (0, 2\pi)$$.

We write $$\sin^2\theta = 1 - \cos^2\theta$$ and $$\cos^2(2\theta) = (2\cos^2\theta - 1)^2$$. Substituting:

$$7\cos^2\theta - 3(1-\cos^2\theta) - 2(2\cos^2\theta-1)^2 = 2$$

$$10\cos^2\theta - 3 - 2(4\cos^4\theta - 4\cos^2\theta + 1) = 2$$

$$10\cos^2\theta - 3 - 8\cos^4\theta + 8\cos^2\theta - 2 = 2$$

$$-8\cos^4\theta + 18\cos^2\theta - 7 = 0$$

$$8\cos^4\theta - 18\cos^2\theta + 7 = 0$$

Let $$u = \cos^2\theta$$. Then $$8u^2 - 18u + 7 = 0$$. By the quadratic formula:

$$u = \frac{18 \pm \sqrt{324 - 224}}{16} = \frac{18 \pm \sqrt{100}}{16} = \frac{18 \pm 10}{16}$$

So $$u = \frac{28}{16} = \frac{7}{4}$$ or $$u = \frac{8}{16} = \frac{1}{2}$$. Since $$\cos^2\theta \leq 1$$, we discard $$u = 7/4$$ and take $$\cos^2\theta = \frac{1}{2}$$, giving $$\cos\theta = \pm\frac{1}{\sqrt{2}}$$.

The solutions in $$(0, 2\pi)$$ are $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$. So the set $$S$$ contains these four values.

For each $$\theta \in S$$, $$\cos^2\theta = \frac{1}{2}$$ and $$\sin^2\theta = \frac{1}{2}$$. Also $$\tan^2\theta = 1$$ and $$\cot^2\theta = 1$$, so $$\tan^2\theta + \cot^2\theta = 2$$.

The quadratic equation becomes $$x^2 - 2(2)x + 6\cdot\frac{1}{2} = 0$$, i.e., $$x^2 - 4x + 3 = 0$$, which factors as $$(x-1)(x-3) = 0$$. The roots are $$x = 1$$ and $$x = 3$$, with sum $$1 + 3 = 4$$.

Since all four values of $$\theta$$ give the same equation (because they all have the same $$\tan^2\theta + \cot^2\theta$$ and $$\sin^2\theta$$), the sum of roots of ALL equations is $$4 \times 4 = 16$$.

Hence, the correct answer is 16.

We need the number of elements in $$A = \{\theta \in S : \tan\theta(1 + \sqrt{5}\tan 2\theta) = \sqrt{5} - \tan 2\theta\}$$ where $$S = \left[-\pi, \frac{\pi}{2}\right) \setminus \left\{-\frac{\pi}{2}, -\frac{\pi}{4}, -\frac{3\pi}{4}, \frac{\pi}{4}\right\}$$.

We rearrange the given equation. Starting from $$\tan\theta + \sqrt{5}\tan\theta\tan 2\theta = \sqrt{5} - \tan 2\theta$$, we move the $$\tan 2\theta$$ term to the left:

$$\tan\theta + \tan 2\theta = \sqrt{5} - \sqrt{5}\tan\theta\tan 2\theta = \sqrt{5}(1 - \tan\theta\tan 2\theta)$$

Dividing both sides by $$(1 - \tan\theta\tan 2\theta)$$ (which must be nonzero for the expression to be valid):

$$\frac{\tan\theta + \tan 2\theta}{1 - \tan\theta\tan 2\theta} = \sqrt{5}$$

The left side is exactly the tangent addition formula $$\tan(\theta + 2\theta) = \tan 3\theta$$. So we need $$\tan 3\theta = \sqrt{5}$$.

Let $$\alpha = \arctan(\sqrt{5})$$, where $$\alpha \in (0, \pi/2)$$ since $$\sqrt{5} > 0$$. The general solution of $$\tan 3\theta = \sqrt{5}$$ is $$3\theta = n\pi + \alpha$$ for integer $$n$$, giving:

$$\theta = \frac{n\pi + \alpha}{3}$$

We need $$\theta \in [-\pi, \pi/2)$$, so $$-\pi \leq \frac{n\pi + \alpha}{3} < \frac{\pi}{2}$$, which gives $$-3\pi \leq n\pi + \alpha < \frac{3\pi}{2}$$.

Since $$0 < \alpha < \pi/2$$: for the left inequality, $$n\pi \geq -3\pi - \alpha > -3\pi - \pi/2 = -7\pi/2$$, so $$n \geq -3$$ (since $$n = -3$$ gives $$-3\pi + \alpha > -3\pi$$, which satisfies $$\geq -3\pi$$). For the right inequality, $$n\pi < 3\pi/2 - \alpha < 3\pi/2$$, so $$n < 3/2$$, meaning $$n \leq 1$$.

This gives $$n \in \{-3, -2, -1, 0, 1\}$$, yielding 5 candidate solutions.

Now we verify none fall in the excluded set $$\{-\pi/2, -\pi/4, -3\pi/4, \pi/4\}$$. For each excluded value $$\theta_0$$, we check $$\tan 3\theta_0$$:

At $$\theta_0 = -\pi/2$$: $$3\theta_0 = -3\pi/2$$, where $$\tan$$ is undefined (not $$\sqrt{5}$$).

At $$\theta_0 = -\pi/4$$: $$\tan(-3\pi/4) = \tan(\pi/4) = 1 \neq \sqrt{5}$$.

At $$\theta_0 = -3\pi/4$$: $$\tan(-9\pi/4) = \tan(-\pi/4) = -1 \neq \sqrt{5}$$.

At $$\theta_0 = \pi/4$$: $$\tan(3\pi/4) = -1 \neq \sqrt{5}$$.

Since none of the excluded points satisfy our equation, no solutions are lost.

We also verify that $$\tan\theta$$ and $$\tan 2\theta$$ are defined at each solution. Since $$\tan 3\theta = \sqrt{5}$$ is finite, $$3\theta \neq \pi/2 + k\pi$$. For $$\tan\theta$$ to be undefined, we need $$\theta = \pi/2 + k\pi$$, which would make $$3\theta = 3\pi/2 + 3k\pi$$, but $$\tan(3\pi/2 + 3k\pi)$$ is undefined, not $$\sqrt{5}$$. Similarly for $$\tan 2\theta$$: if $$2\theta = \pi/2 + k\pi$$, then $$\theta = \pi/4 + k\pi/2$$, giving $$3\theta = 3\pi/4 + 3k\pi/2$$, and $$\tan(3\pi/4) = -1 \neq \sqrt{5}$$. So all 5 solutions have both $$\tan\theta$$ and $$\tan 2\theta$$ well-defined.

Hence, the correct answer is 5.

We are given that $$\log_{10} \sin x + \log_{10} \cos x = -1$$ for $$x \in \left(0, \frac{\pi}{2}\right)$$. This means $$\log_{10}(\sin x \cos x) = -1$$, so $$\sin x \cos x = 10^{-1} = \frac{1}{10}$$.

We know that $$\sin x \cos x = \frac{\sin 2x}{2}$$, so $$\frac{\sin 2x}{2} = \frac{1}{10}$$, giving $$\sin 2x = \frac{1}{5}$$.

Now we need to find $$(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2 \times \frac{1}{10} = 1 + \frac{1}{5} = \frac{6}{5}$$.

Since $$x \in \left(0, \frac{\pi}{2}\right)$$, both $$\sin x$$ and $$\cos x$$ are positive, so $$\sin x + \cos x = \sqrt{\frac{6}{5}}$$.

Now we use the second equation: $$\log_{10}(\sin x + \cos x) = \frac{1}{2}(\log_{10} n - 1)$$.

The left side is $$\log_{10}\sqrt{\frac{6}{5}} = \frac{1}{2}\log_{10}\frac{6}{5}$$.

So $$\frac{1}{2}\log_{10}\frac{6}{5} = \frac{1}{2}(\log_{10} n - 1)$$.

Multiplying both sides by 2: $$\log_{10}\frac{6}{5} = \log_{10} n - 1$$.

This gives $$\log_{10} n = 1 + \log_{10}\frac{6}{5} = \log_{10} 10 + \log_{10}\frac{6}{5} = \log_{10}\left(10 \times \frac{6}{5}\right) = \log_{10} 12$$.

Therefore $$n = 12$$, and the correct answer is option (2).

We start with the given transcendental equation

$$32^{\tan^2x}+32^{\sec^2x}=81,\qquad 0\le x\le\frac{\pi}{4}.$$

First we recall the Pythagorean identity for tangent and secant:

$$\sec^2x\,=\,1+\tan^2x.$$

Let us put

$$t=\tan^2x.$$

Because $$0\le x\le\frac{\pi}{4}$$, we have $$0\le \tan x\le1$$; hence

$$0\le t\le1.$$

Using $$\sec^2x=1+t$$ in the original equation gives

$$32^{t}+32^{\,1+t}=81.$$

By the property of exponents $$a^{m+n}=a^{m}\,a^{n}$$, we write

$$32^{\,1+t}=32^{1}\cdot32^{t}=32\;32^{t}.$$

Substituting this result back, the equation becomes

$$32^{t}+32\;32^{t}=81.$$

Taking $$32^{t}$$ common we obtain

$$32^{t}\bigl(1+32\bigr)=81.$$

Since $$1+32=33$$, we have

$$32^{t}\,.\,33=81.$$

Dividing both sides by $$33$$:

$$32^{t}=\frac{81}{33}=\frac{27}{11}.$$

Now $$32=2^{5}$$. Hence

$$32^{t}=(2^{5})^{t}=2^{5t}=\frac{27}{11}.$$

Taking base-$$2$$ logarithms on both sides (formula: if $$a^b=c$$, then $$b=\log_{a}c$$):

$$5t=\log_{2}\!\Bigl(\tfrac{27}{11}\Bigr).$$

Therefore

$$t=\frac{1}{5}\,\log_{2}\!\Bigl(\tfrac{27}{11}\Bigr).$$

Because $$\tfrac{27}{11}\approx2.4545$$ and $$\log_{2}(2.4545)\approx1.296$$, we get a numerical value

$$t\approx\frac{1.296}{5}\approx0.259.$$

Remember $$t=\tan^{2}x$$. So

$$\tan^{2}x\approx0.259.$$

Taking the positive square root (only the positive root is relevant here since $$\tan x\ge0$$ on the interval $$0\le x\le\frac{\pi}{4}$$):

$$\tan x\approx\sqrt{0.259}\approx0.509.$$

Because the tangent function is strictly increasing on $$\bigl[0,\tfrac{\pi}{4}\bigr]$$, the equation $$\tan x\approx0.509$$ possesses exactly one solution in that interval. Consequently the original equation has exactly one admissible $$x$$ between $$0$$ and $$\tfrac{\pi}{4}$$.

Hence, the correct answer is Option C.

We are given $$15\sin^4\alpha + 10\cos^4\alpha = 6$$. Let $$\sin^2\alpha = s$$ and $$\cos^2\alpha = 1 - s$$, so the equation becomes $$15s^2 + 10(1-s)^2 = 6$$, which expands to $$15s^2 + 10 - 20s + 10s^2 = 6$$, giving $$25s^2 - 20s + 4 = 0$$. This factors as $$(5s - 2)^2 = 0$$, so $$s = \frac{2}{5}$$, meaning $$\sin^2\alpha = \frac{2}{5}$$ and $$\cos^2\alpha = \frac{3}{5}$$.

Now we compute $$27\sec^6\alpha + 8\csc^6\alpha = \frac{27}{\cos^6\alpha} + \frac{8}{\sin^6\alpha} = \frac{27}{(3/5)^3} + \frac{8}{(2/5)^3} = \frac{27 \cdot 125}{27} + \frac{8 \cdot 125}{8} = 125 + 125 = 250$$.

We are given that $$e^{(\cos^2 x + \cos^4 x + \cos^6 x + \ldots \infty) \ln 2}$$ satisfies the equation $$t^2 - 9t + 8 = 0$$.

The infinite geometric series $$\cos^2 x + \cos^4 x + \cos^6 x + \ldots$$ has first term $$\cos^2 x$$ and common ratio $$\cos^2 x$$. Its sum is $$\frac{\cos^2 x}{1 - \cos^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x$$.

So the expression becomes $$e^{\cot^2 x \cdot \ln 2} = 2^{\cot^2 x}$$.

Now, $$t^2 - 9t + 8 = 0$$ gives $$(t - 1)(t - 8) = 0$$, so $$t = 1$$ or $$t = 8$$.

If $$2^{\cot^2 x} = 1$$, then $$\cot^2 x = 0$$, so $$x = \frac{\pi}{2}$$.

If $$2^{\cot^2 x} = 8 = 2^3$$, then $$\cot^2 x = 3$$, so $$\cot x = \sqrt{3}$$ (since $$0 < x < \frac{\pi}{2}$$), giving $$x = \frac{\pi}{6}$$.

Now we evaluate $$\frac{2\sin x}{\sin x + \sqrt{3}\cos x}$$ at $$x = \frac{\pi}{6}$$:

$$\frac{2 \cdot \frac{1}{2}}{\frac{1}{2} + \sqrt{3} \cdot \frac{\sqrt{3}}{2}} = \frac{1}{\frac{1}{2} + \frac{3}{2}} = \frac{1}{2}$$.

At $$x = \frac{\pi}{2}$$: $$\frac{2 \cdot 1}{1 + 0} = 2$$, which is not among the options.

Hence, the correct answer is Option B.

We have that $$\tan\left(\frac{\pi}{9}\right),\,x,\,\tan\left(\frac{7\pi}{18}\right)$$ form an arithmetic progression, so by the definition of an A.P. the middle term is the average of the other two. Hence

$$2x=\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{7\pi}{18}\right).$$

Dividing by 2 we obtain

$$x=\frac{\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{7\pi}{18}\right)}{2}.$$

In the same way, the numbers $$\tan\left(\frac{\pi}{9}\right),\,y,\,\tan\left(\frac{5\pi}{18}\right)$$ are in arithmetic progression, so

$$2y=\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{5\pi}{18}\right)$$

and therefore

$$y=\frac{\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{5\pi}{18}\right)}{2}.$$

To handle these expressions we set

$$t=\tan\left(\frac{\pi}{9}\right)=\tan 20^{\circ}.$$

The angle $$\frac{7\pi}{18}=70^{\circ}$$ is the complement of $$20^{\circ}$$, so

$$\tan\left(\frac{7\pi}{18}\right)=\tan 70^{\circ}=\cot 20^{\circ}=\frac1{\tan 20^{\circ}}=\frac1t.$$

Thus

$$x=\frac{t+\dfrac1t}{2}.$$

Next we need $$\tan\left(\frac{5\pi}{18}\right)=\tan 50^{\circ}.$$ Let us denote

$$p=\tan 50^{\circ}.$$

Because $$70^{\circ}=50^{\circ}+20^{\circ},$$ we can apply the tangent addition formula

$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\,\tan B}.$$

Taking $$A=50^{\circ}$$ and $$B=20^{\circ},$$ we have

$$\tan 70^{\circ}=\frac{\tan 50^{\circ}+\tan 20^{\circ}}{1-\tan 50^{\circ}\tan 20^{\circ}}.$$

Substituting the symbols just introduced, this becomes

$$\frac1t=\frac{p+t}{1-pt}.$$

Cross-multiplying gives

$$t(p+t)=1-pt.$$

Expanding and bringing all terms to one side we get

$$tp+t^{2}=1-pt,$$

$$tp+pt+t^{2}=1,$$

$$2pt+t^{2}=1.$$

Solving for $$p$$ we find

$$p=\frac{1-t^{2}}{2t}.$$

Now we can write $$y$$ solely in terms of $$t$$:

$$y=\frac{t+p}{2}=\frac{t+\dfrac{1-t^{2}}{2t}}{2}.$$

It is more convenient to work directly with $$2y$$, because $$|x-2y|$$ appears in the question. We have from earlier

$$2y=t+p=t+\frac{1-t^{2}}{2t}.$$

Next we compute the required difference. Starting from the expression for $$x$$ we write

$$x-2y=\frac{t+\dfrac1t}{2}-\left(t+\frac{1-t^{2}}{2t}\right).$$

To combine terms we bring everything to a common denominator $$2t$$:

$$x-2y=\frac{t^{2}+1}{2t}-\frac{2t^{2}+1-t^{2}}{2t}.$$

Simplifying the numerator of the second fraction gives

$$2t^{2}+1-t^{2}=t^{2}+1,$$

so that

$$x-2y=\frac{t^{2}+1}{2t}-\frac{t^{2}+1}{2t}=0.$$

Hence

$$|x-2y|=|0|=0.$$

Hence, the correct answer is Option C.

Let us denote $$\theta = 18^\circ$$ and write the number we are interested in as $$x = \csc 18^\circ.$$ Our task is to find an equation with integer coefficients that has this value of $$x$$ as a root.

Because trigonometric identities for $$18^\circ$$ are most easily obtained through $$36^\circ$$, we begin by finding $$\cos 36^\circ.$$ Observe that $$5 \times 36^\circ = 180^\circ,$$ so $$\cos 5(36^\circ) = \cos 180^\circ = -1.$$

The standard multiple-angle formula for cosine is

$$\cos 5\theta = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta.$$

Putting $$\theta = 36^\circ$$ and writing $$c = \cos 36^\circ,$$ we obtain

$$16c^5 - 20c^3 + 5c + 1 = 0.$$

This quintic has a quadratic factor which can be found by inspection (or long division); that factor is $$4c^2 - 2c - 1.$$ Hence

$$4c^2 - 2c - 1 = 0.$$

Solving the quadratic with the quadratic formula,

$$c = \dfrac{2 \pm \sqrt{4 + 16}}{8} = \dfrac{2 \pm 4\sqrt5}{8} = \dfrac{1 \pm \sqrt5}{4}.$$

Since $$\cos 36^\circ$$ is positive, we choose the positive root:

$$\cos 36^\circ = \dfrac{1 + \sqrt5}{4}.$$

Now we convert this to $$\sin 18^\circ$$ using the half-angle relation

$$\sin^2 18^\circ = \dfrac{1 - \cos 36^\circ}{2}.$$

Substituting the value of $$\cos 36^\circ,$$ we get

$$\sin^2 18^\circ = \dfrac{1 - \dfrac{1 + \sqrt5}{4}}{2} = \dfrac{\dfrac{4 - 1 - \sqrt5}{4}}{2} = \dfrac{3 - \sqrt5}{8}.$$

Taking the positive square root (because $$\sin 18^\circ$$ is positive),

$$\sin 18^\circ = \sqrt{\dfrac{3 - \sqrt5}{8}}.$$

A more familiar exact form arises by rationalising:

$$\sin 18^\circ = \dfrac{\sqrt5 - 1}{4}.$$

We can check this quickly, because

$$\left(\dfrac{\sqrt5 - 1}{4}\right)^2 = \dfrac{5 + 1 - 2\sqrt5}{16} = \dfrac{6 - 2\sqrt5}{16} = \dfrac{3 - \sqrt5}{8},$$

matching the value found earlier. Now we are ready to obtain $$\csc 18^\circ.$$ Since $$\csc 18^\circ = \dfrac{1}{\sin 18^\circ},$$ we have

$$x = \csc 18^\circ = \dfrac{1}{\dfrac{\sqrt5 - 1}{4}} = \dfrac{4}{\sqrt5 - 1}.$$

To eliminate the radical from the denominator we multiply numerator and denominator by $$\sqrt5 + 1$$:

$$x = \dfrac{4(\sqrt5 + 1)}{(\sqrt5 - 1)(\sqrt5 + 1)} = \dfrac{4(\sqrt5 + 1)}{5 - 1} = \dfrac{4(\sqrt5 + 1)}{4} = \sqrt5 + 1.$$

Therefore

$$x = \sqrt5 + 1.$$

Let us now examine which of the given quadratic equations is satisfied by $$x.$$ We compute the left-hand side of each candidate equation, starting with

$$x^2 - 2x - 4.$$

First, we find $$x^2$$ and $$2x$$:

$$x^2 = (\sqrt5 + 1)^2 = 5 + 1 + 2\sqrt5 = 6 + 2\sqrt5,$$

$$2x = 2(\sqrt5 + 1) = 2\sqrt5 + 2.$$

Now substitute these into the expression:

$$x^2 - 2x - 4 = (6 + 2\sqrt5) - (2\sqrt5 + 2) - 4.$$

Simplifying term by term, the $$2\sqrt5$$ terms cancel:

$$= 6 + 2\sqrt5 - 2\sqrt5 - 2 - 4 = 6 - 2 - 4 = 0.$$

Thus $$x = \csc 18^\circ$$ satisfies

$$x^2 - 2x - 4 = 0.$$

Checking the other options is unnecessary, because we have already found an equation with integer coefficients that our number satisfies, and only one such equation appears in the list.

Hence, the correct answer is Option A.

We need to find the number of solutions of $$\sin^7 x + \cos^7 x = 1$$ in $$[0, 4\pi]$$.

We know that for all real $$x$$, $$|\sin x| \leq 1$$ and $$|\cos x| \leq 1$$. Since the exponent 7 is odd, $$\sin^7 x$$ has the same sign as $$\sin x$$. For $$|t| \leq 1$$, we have $$|t|^7 \leq t^2$$ (since $$|t|^5 \leq 1$$).

We can write: $$\sin^7 x + \cos^7 x = 1 = \sin^2 x + \cos^2 x$$, which gives $$(\sin^2 x - \sin^7 x) + (\cos^2 x - \cos^7 x) = 0$$, i.e., $$\sin^2 x(1 - \sin^5 x) + \cos^2 x(1 - \cos^5 x) = 0$$.

For any real $$x$$, if $$\sin x \geq 0$$ then $$\sin^5 x \leq 1$$ so the first term is $$\geq 0$$. If $$\sin x < 0$$ then $$\sin^5 x < 0$$ so $$1 - \sin^5 x > 1$$ and the first term is still $$> 0$$ (unless $$\sin x = 0$$). The same reasoning applies to the cosine term. Therefore each term is non-negative, and their sum can only be zero if each term is individually zero.

$$\sin^2 x(1 - \sin^5 x) = 0$$ gives $$\sin x = 0$$ or $$\sin x = 1$$.

$$\cos^2 x(1 - \cos^5 x) = 0$$ gives $$\cos x = 0$$ or $$\cos x = 1$$.

Case 1: $$\sin x = 0$$ and $$\cos x = 1$$, i.e., $$x = 2k\pi$$. In $$[0, 4\pi]$$: $$x = 0, 2\pi, 4\pi$$ — giving 3 solutions.

Case 2: $$\sin x = 1$$ and $$\cos x = 0$$, i.e., $$x = \frac{\pi}{2} + 2k\pi$$. In $$[0, 4\pi]$$: $$x = \frac{\pi}{2}, \frac{5\pi}{2}$$ — giving 2 solutions.

(The combinations $$\sin x = 0, \cos x = 0$$ and $$\sin x = 1, \cos x = 1$$ are both impossible.)

Total number of solutions = $$3 + 2 = 5$$.

The answer is $$5$$, which is Option C.

We start from the given trigonometric equation

$$$\sin x+\sin 2x+\sin 3x+\sin 4x=0\;,\qquad x\in[0,\,2\pi].$$$

In order to combine the terms pair-wise, we recall the sum-to-product identity

$$\sin A+\sin B \;=\; 2\sin\!\left(\dfrac{A+B}{2}\right)\cos\!\left(\dfrac{A-B}{2}\right).$$

Applying this once to the first and last terms and once to the two middle terms we obtain

$$$ \begin{aligned} \sin x+\sin 4x &= 2\sin\!\left(\dfrac{x+4x}{2}\right)\cos\!\left(\dfrac{x-4x}{2}\right) = 2\sin\!\left(\dfrac{5x}{2}\right)\cos\!\left(-\dfrac{3x}{2}\right) = 2\sin\!\left(\dfrac{5x}{2}\right)\cos\!\left(\dfrac{3x}{2}\right),\\[6pt] \sin 2x+\sin 3x &= 2\sin\!\left(\dfrac{2x+3x}{2}\right)\cos\!\left(\dfrac{2x-3x}{2}\right) = 2\sin\!\left(\dfrac{5x}{2}\right)\cos\!\left(-\dfrac{x}{2}\right) = 2\sin\!\left(\dfrac{5x}{2}\right)\cos\!\left(\dfrac{x}{2}\right). \end{aligned} $$$

Adding these two results we get

$$$ \sin x+\sin 2x+\sin 3x+\sin 4x = 2\sin\!\left(\dfrac{5x}{2}\right)\Bigl[\cos\!\left(\dfrac{3x}{2}\right)+\cos\!\left(\dfrac{x}{2}\right)\Bigr]. $$$

Hence the original equation becomes

$$2\sin\!\left(\dfrac{5x}{2}\right)\Bigl[\cos\!\left(\dfrac{3x}{2}\right)+\cos\!\left(\dfrac{x}{2}\right)\Bigr]=0.$$ So we must have

$$$\sin\!\left(\dfrac{5x}{2}\right)=0 \quad\text{or}\quad \cos\!\left(\dfrac{3x}{2}\right)+\cos\!\left(\dfrac{x}{2}\right)=0.$$$ We analyse the two cases separately.

Case 1. $$\sin\!\left(\dfrac{5x}{2}\right)=0.$$
The sine of any angle is zero when the angle equals an integer multiple of $$\pi$$, that is, when $$\dfrac{5x}{2}=n\pi,\qquad n\in\mathbb Z.$$ Solving for $$x$$ gives $$x=\dfrac{2n\pi}{5}.$$ Because $$x$$ must lie in $$[0,2\pi]$$ we list all permissible integers $$n$$:

$$$ \begin{aligned} n=0 &\;\Rightarrow\; x=0,\\ n=1 &\;\Rightarrow\; x=\dfrac{2\pi}{5},\\ n=2 &\;\Rightarrow\; x=\dfrac{4\pi}{5},\\ n=3 &\;\Rightarrow\; x=\dfrac{6\pi}{5},\\ n=4 &\;\Rightarrow\; x=\dfrac{8\pi}{5},\\ n=5 &\;\Rightarrow\; x=2\pi. \end{aligned} $$$

The sum of these six values equals

$$$ 0+\dfrac{2\pi}{5}+\dfrac{4\pi}{5}+\dfrac{6\pi}{5}+\dfrac{8\pi}{5}+2\pi =\dfrac{20\pi}{5}+2\pi =4\pi+2\pi =6\pi. $$$

Case 2. $$\cos\!\left(\dfrac{3x}{2}\right)+\cos\!\left(\dfrac{x}{2}\right)=0.$$
For the cosine sum we employ the identity

$$\cos A+\cos B = 2\cos\!\left(\dfrac{A+B}{2}\right)\cos\!\left(\dfrac{A-B}{2}\right).$$

Putting $$A=\dfrac{3x}{2}$$ and $$B=\dfrac{x}{2}$$ we get

$$\cos\!\left(\dfrac{3x}{2}\right)+\cos\!\left(\dfrac{x}{2}\right) =2\cos x\,\cos\!\left(\dfrac{x}{2}\right).$$

Thus the second factorised equation is

$$2\cos x\,\cos\!\left(\dfrac{x}{2}\right)=0,$$ which leads to

$$\cos x = 0 \quad\text{or}\quad \cos\!\left(\dfrac{x}{2}\right)=0.$$ We treat these two possibilities one by one.

(i) $$\cos x=0.$$
The cosine vanishes at odd multiples of $$\frac{\pi}{2}$$, so

$$x=\dfrac{\pi}{2}+n\pi,\qquad n\in\mathbb Z.$$

Inside the interval $$[0,2\pi]$$ this gives

$$x=\dfrac{\pi}{2},\; \dfrac{3\pi}{2}.$$

Their sum is

$$\dfrac{\pi}{2}+\dfrac{3\pi}{2}=2\pi.$$

(ii) $$\cos\!\left(\dfrac{x}{2}\right)=0.$$
Again, cosine is zero at odd multiples of $$\frac{\pi}{2}$$, so

$$\dfrac{x}{2}=\dfrac{\pi}{2}+m\pi,\qquad m\in\mathbb Z$$ which gives $$x=\pi+2m\pi.$$

Within $$[0,2\pi]$$ we obtain only

$$x=\pi,$$ whose contribution to the sum of roots is simply $$\pi.$$

Combining the results from both cases, we list all distinct solutions

$$$\Bigl\{\,0,\ \dfrac{2\pi}{5},\ \dfrac{4\pi}{5},\ \dfrac{6\pi}{5},\ \dfrac{8\pi}{5},\ 2\pi,\ \dfrac{\pi}{2},\ \dfrac{3\pi}{2},\ \pi\Bigr\}.$$$ There are nine in total, and their sum equals

$$ 6\pi\;+\;2\pi\;+\;\pi =\;9\pi. $$

Hence, the correct answer is Option D.

We have to evaluate

$$$2\sin\!\left(\frac{\pi}{8}\right)\sin\!\left(\frac{2\pi}{8}\right)\sin\!\left(\frac{3\pi}{8}\right)\sin\!\left(\frac{5\pi}{8}\right)\sin\!\left(\frac{6\pi}{8}\right)\sin\!\left(\frac{7\pi}{8}\right).$$$

First we notice the trigonometric fact $$\sin(\theta)=\sin(\pi-\theta).$$ Using this fact, three of the six sines can be rewritten:

$$\sin\!\left(\frac{5\pi}{8}\right)=\sin\!\left(\pi-\frac{5\pi}{8}\right)=\sin\!\left(\frac{3\pi}{8}\right),$$ $$\sin\!\left(\frac{6\pi}{8}\right)=\sin\!\left(\pi-\frac{6\pi}{8}\right)=\sin\!\left(\frac{2\pi}{8}\right),$$ $$\sin\!\left(\frac{7\pi}{8}\right)=\sin\!\left(\pi-\frac{7\pi}{8}\right)=\sin\!\left(\frac{\pi}{8}\right).$$

So the entire product pairs up neatly:

$$\sin\!\left(\frac{\pi}{8}\right)\sin\!\left(\frac{5\pi}{8}\right)=\sin^2\!\left(\frac{\pi}{8}\right),$$ $$\sin\!\left(\frac{2\pi}{8}\right)\sin\!\left(\frac{6\pi}{8}\right)=\sin^2\!\left(\frac{\pi}{4}\right),$$ $$\sin\!\left(\frac{3\pi}{8}\right)\sin\!\left(\frac{7\pi}{8}\right)=\sin^2\!\left(\frac{3\pi}{8}\right).$$

Hence the whole expression becomes

$$$2\;\times\;\sin^2\!\left(\frac{\pi}{8}\right)\;\times\;\sin^2\!\left(\frac{\pi}{4}\right)\;\times\;\sin^2\!\left(\frac{3\pi}{8}\right).$$$

We already know $$\sin\!\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2},$$ so

$$\sin^2\!\left(\frac{\pi}{4}\right)=\left(\frac{\sqrt{2}}{2}\right)^2=\frac12.$$

Now we compute $$\sin\!\left(\frac{\pi}{8}\right)$$ and $$\sin\!\left(\frac{3\pi}{8}\right)$$ using the half-angle formula $$\sin^2\!\left(\frac{\alpha}{2}\right)=\frac{1-\cos\alpha}{2}.$$

Taking $$\alpha=\frac{\pi}{4},$$ we get

$$\sin^2\!\left(\frac{\pi}{8}\right)=\frac{1-\cos\!\left(\frac{\pi}{4}\right)}{2}=\frac{1-\frac{\sqrt{2}}{2}}{2}.$$

Taking $$\alpha=\frac{3\pi}{4},$$ we get

$$$\sin^2\!\left(\frac{3\pi}{8}\right)=\frac{1-\cos\!\left(\frac{3\pi}{4}\right)}{2}=\frac{1-\left(-\frac{\sqrt{2}}{2}\right)}{2}=\frac{1+\frac{\sqrt{2}}{2}}{2}.$$$

Let us set

$$$A=\sin^2\!\left(\frac{\pi}{8}\right)=\frac{1-\frac{\sqrt{2}}{2}}{2}, \qquad B=\sin^2\!\left(\frac{3\pi}{8}\right)=\frac{1+\frac{\sqrt{2}}{2}}{2}.$$$

The product of these two is

$$$AB=\frac{1-\frac{\sqrt{2}}{2}}{2}\;\times\;\frac{1+\frac{\sqrt{2}}{2}}{2}=\frac{\left(1-\frac{\sqrt{2}}{2}\right)\!\left(1+\frac{\sqrt{2}}{2}\right)}{4}.$$$

Using $$\left(x-y\right)\left(x+y\right)=x^2-y^2,$$ take $$x=1,\;y=\frac{\sqrt{2}}{2}.$$ Thus the numerator becomes

$$1^2-\left(\frac{\sqrt{2}}{2}\right)^2=1-\frac{2}{4}=1-\frac12=\frac12.$$

Therefore

$$AB=\frac{\frac12}{4}=\frac18.$$

Now multiply by $$\sin^2\!\left(\frac{\pi}{4}\right)=\frac12:$$

$$A\,B\,\sin^2\!\left(\frac{\pi}{4}\right)=\frac18\times\frac12=\frac1{16}.$$

This is the value of the six-sine product without the leading 2. Finally, include that factor 2:

$$2\times\frac1{16}=\frac18.$$

Hence, the correct answer is Option B.

We are given $$\cos x + \cos y - \cos(x + y) = \frac{3}{2}$$ with $$0 < x, y < \pi$$.

Using the identities $$\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}$$ and $$\cos(x+y) = 2\cos^2\frac{x+y}{2} - 1$$, we substitute to get $$2\cos\frac{x+y}{2}\cos\frac{x-y}{2} - 2\cos^2\frac{x+y}{2} + 1 = \frac{3}{2}$$.

Let $$u = \cos\frac{x+y}{2}$$ and $$v = \cos\frac{x-y}{2}$$. Then $$2uv - 2u^2 = \frac{1}{2}$$, or equivalently $$4u^2 - 4uv + 1 = 0$$.

Treating this as a quadratic in $$u$$: $$4u^2 - 4vu + 1 = 0$$. The discriminant must be non-negative: $$16v^2 - 16 \geq 0$$, so $$v^2 \geq 1$$. Since $$v = \cos\frac{x-y}{2}$$ and $$|v| \leq 1$$, we need $$v = \pm 1$$. Given the constraint $$0 < x, y < \pi$$, we have $$\left|\frac{x-y}{2}\right| < \frac{\pi}{2}$$, so $$v > 0$$, giving $$v = 1$$, which means $$x = y$$.

With $$x = y$$, the equation becomes $$2\cos x - \cos 2x = \frac{3}{2}$$, i.e., $$2\cos x - (2\cos^2 x - 1) = \frac{3}{2}$$, which simplifies to $$2\cos^2 x - 2\cos x + \frac{1}{2} = 0$$, or $$4\cos^2 x - 4\cos x + 1 = 0$$.

This gives $$(2\cos x - 1)^2 = 0$$, so $$\cos x = \frac{1}{2}$$, meaning $$x = \frac{\pi}{3}$$. Therefore $$x = y = \frac{\pi}{3}$$.

Thus, $$\sin x + \cos y = \sin\frac{\pi}{3} + \cos\frac{\pi}{3} = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1 + \sqrt{3}}{2}$$.

We have been told that $$\sin\theta+\cos\theta=\dfrac12.$$

First, we square both sides. Using the identity $$(a+b)^2=a^2+b^2+2ab,$$ we obtain

$$\left(\sin\theta+\cos\theta\right)^2=\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta.$$

Because $$\sin^2\theta+\cos^2\theta=1,$$ the right-hand side becomes $$1+2\sin\theta\cos\theta.$$ Therefore

$$\left(\sin\theta+\cos\theta\right)^2=1+2\sin\theta\cos\theta.$$

Substituting the given value $$\sin\theta+\cos\theta=\dfrac12,$$ we get

$$\left(\dfrac12\right)^2=1+2\sin\theta\cos\theta,$$

$$\dfrac14=1+2\sin\theta\cos\theta.$$

So,

$$2\sin\theta\cos\theta=\dfrac14-1=-\dfrac34.$$

The double-angle formula $$\sin2\theta=2\sin\theta\cos\theta$$ now gives

$$\sin2\theta=-\dfrac34.$$

Next, we need $$\cos4\theta.$$ Using the double-angle identity $$\cos2A=1-2\sin^2A,$$ we write

$$\cos4\theta=\cos\bigl(2\cdot2\theta\bigr)=1-2\sin^22\theta.$$

Since $$\sin2\theta=-\dfrac34,$$ we have

$$\sin^22\theta=\left(-\dfrac34\right)^2=\dfrac9{16}.$$

Thus,

$$\cos4\theta=1-2\left(\dfrac9{16}\right)=1-\dfrac{18}{16}=\dfrac{16-18}{16}=-\dfrac18.$$

Now we compute $$\sin6\theta.$$ The angle-addition formula $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ gives

$$\sin6\theta=\sin\bigl(4\theta+2\theta\bigr)=\sin4\theta\cos2\theta+\cos4\theta\sin2\theta.$$

To use this, we first find $$\cos2\theta.$$ From $$\sin^22\theta+\cos^22\theta=1$$ we get

$$\cos^22\theta=1-\sin^22\theta=1-\dfrac9{16}=\dfrac7{16},$$

$$\cos2\theta=\pm\dfrac{\sqrt7}{4}.$$

We will keep the sign undetermined because the final numerical value will turn out the same in either case.

Using the double-angle identity $$\sin4\theta=2\sin2\theta\cos2\theta,$$ we obtain

$$\sin4\theta=2\left(-\dfrac34\right)\left(\pm\dfrac{\sqrt7}{4}\right)=\mp\dfrac{3\sqrt7}{8}.$$

Now substitute into $$\sin6\theta=\sin4\theta\cos2\theta+\cos4\theta\sin2\theta:$$

$$\sin6\theta=\left(\mp\dfrac{3\sqrt7}{8}\right)\left(\pm\dfrac{\sqrt7}{4}\right)+\left(-\dfrac18\right)\left(-\dfrac34\right).$$

The product of the first pair of signs is negative, so the term simplifies to $$-\dfrac{3\cdot7}{32}=-\dfrac{21}{32}.$$ The second term equals $$\dfrac{3}{32}.$$ Therefore

$$\sin6\theta=-\dfrac{21}{32}+\dfrac{3}{32}=-\dfrac{18}{32}=-\dfrac{9}{16}.$$

We are finally ready to evaluate the required expression:

$$16\left(\sin2\theta+\cos4\theta+\sin6\theta\right)=16\left(-\dfrac34-\dfrac18-\dfrac{9}{16}\right).$$

Convert everything to a common denominator $$16:$$

$$-\dfrac34=-\dfrac{12}{16},\quad -\dfrac18=-\dfrac{2}{16},\quad -\dfrac{9}{16}=-\dfrac{9}{16}.$$

Adding these gives

$$-\dfrac{12}{16}-\dfrac{2}{16}-\dfrac{9}{16}=-\dfrac{23}{16}.$$

Multiplying by $$16$$, we obtain

$$16\left(-\dfrac{23}{16}\right)=-23.$$

Hence, the correct answer is Option C.

We need to find the number of solutions of $$x + 2\tan x = \frac{\pi}{2}$$ in $$[0, 2\pi]$$.

Rearranging, we get $$2\tan x = \frac{\pi}{2} - x$$, which gives $$\tan x = \frac{1}{2}\left(\frac{\pi}{2} - x\right)$$.

Let us define $$f(x) = \tan x$$ and $$g(x) = \frac{1}{2}\left(\frac{\pi}{2} - x\right)$$. The solutions are the intersection points of these two curves in $$[0, 2\pi]$$.

The function $$g(x)$$ is a straight line with slope $$-\frac{1}{2}$$ and $$y$$-intercept $$\frac{\pi}{4}$$. At $$x = 0$$, $$g(0) = \frac{\pi}{4} \approx 0.785$$. At $$x = \frac{\pi}{2}$$, $$g\left(\frac{\pi}{2}\right) = 0$$. At $$x = \pi$$, $$g(\pi) = -\frac{\pi}{4}$$. At $$x = 2\pi$$, $$g(2\pi) = -\frac{3\pi}{4}$$.

The function $$\tan x$$ has vertical asymptotes at $$x = \frac{\pi}{2}$$ and $$x = \frac{3\pi}{2}$$, and is continuous on the intervals $$\left[0, \frac{\pi}{2}\right)$$, $$\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$$, and $$\left(\frac{3\pi}{2}, 2\pi\right]$$.

In the interval $$\left[0, \frac{\pi}{2}\right)$$: $$\tan x$$ increases from $$0$$ to $$+\infty$$, while $$g(x)$$ decreases from $$\frac{\pi}{4}$$ to $$0$$. At $$x = 0$$, $$\tan 0 = 0 < \frac{\pi}{4} = g(0)$$. As $$x \to \frac{\pi}{2}^-$$, $$\tan x \to +\infty$$ while $$g(x) \to 0$$. So $$\tan x$$ crosses $$g(x)$$ exactly once in this interval.

In the interval $$\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$$: $$\tan x$$ goes from $$-\infty$$ to $$+\infty$$. The line $$g(x)$$ is continuous on this interval, taking finite values. Since $$\tan x$$ is continuous and strictly increasing on this interval, going from $$-\infty$$ to $$+\infty$$, it crosses the line $$g(x)$$ exactly once.

In the interval $$\left(\frac{3\pi}{2}, 2\pi\right]$$: $$\tan x$$ goes from $$-\infty$$ to $$0$$. At $$x = \frac{3\pi}{2}$$, $$g\left(\frac{3\pi}{2}\right) = \frac{1}{2}\left(\frac{\pi}{2} - \frac{3\pi}{2}\right) = -\frac{\pi}{2} \approx -1.57$$. At $$x = 2\pi$$, $$g(2\pi) = -\frac{3\pi}{4} \approx -2.36$$ and $$\tan(2\pi) = 0$$. Since $$\tan x$$ comes from $$-\infty$$ and increases to $$0$$, while $$g(x)$$ takes finite negative values (between $$-\frac{\pi}{2}$$ and $$-\frac{3\pi}{4}$$), the curves cross exactly once.

Therefore, the total number of solutions is $$3$$, which is Option A.

We need to find all $$\theta \in [0, 2\pi]$$ for which $$\sin 2\theta + \tan 2\theta > 0$$.

First, note that $$\tan 2\theta$$ is undefined when $$\cos 2\theta = 0$$, i.e., at $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$. We exclude these values.

For all other $$\theta$$, we combine the terms: $$\sin 2\theta + \tan 2\theta = \sin 2\theta + \frac{\sin 2\theta}{\cos 2\theta} = \sin 2\theta \cdot \frac{\cos 2\theta + 1}{\cos 2\theta}$$.

Using the double angle identity $$\cos 2\theta + 1 = 2\cos^2\theta$$, the expression becomes $$\frac{2\sin 2\theta \cos^2\theta}{\cos 2\theta}$$.

Now $$\cos^2\theta = 0$$ only at $$\theta = \frac{\pi}{2}, \frac{3\pi}{2}$$. At these points, $$\sin 2\theta = \sin\pi = 0$$ and $$\sin 3\pi = 0$$ respectively, and $$\tan 2\theta = \tan\pi = 0$$ and $$\tan 3\pi = 0$$ respectively. So $$\sin 2\theta + \tan 2\theta = 0$$ at these points, and the inequality is not satisfied. We exclude them.

For all remaining $$\theta$$, we have $$\cos^2\theta > 0$$, so $$2\cos^2\theta > 0$$. Therefore the sign of the expression depends only on $$\frac{\sin 2\theta}{\cos 2\theta} = \tan 2\theta$$.

The inequality reduces to $$\tan 2\theta > 0$$. The tangent function is positive in the first and third quadrants: $$\tan 2\theta > 0$$ when $$2\theta \in \left(0, \frac{\pi}{2}\right) \cup \left(\pi, \frac{3\pi}{2}\right) \cup \left(2\pi, \frac{5\pi}{2}\right) \cup \left(3\pi, \frac{7\pi}{2}\right)$$.

Dividing all bounds by 2: $$\theta \in \left(0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{4}\right) \cup \left(\pi, \frac{5\pi}{4}\right) \cup \left(\frac{3\pi}{2}, \frac{7\pi}{4}\right)$$.

Therefore, the solution set is $$\left(0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{4}\right) \cup \left(\pi, \frac{5\pi}{4}\right) \cup \left(\frac{3\pi}{2}, \frac{7\pi}{4}\right)$$.

We have to solve the trigonometric-algebraic equation

$$2\cos x\cdot4\sin\frac{\pi}{4}+x\sin\frac{\pi}{4}-x-1=1,\qquad 0\le x\le\pi.$$

First we evaluate the constant trigonometric numbers that appear. We know that

$$\sin\frac{\pi}{4}=\frac{\sqrt2}{2}.$$

Substituting this value in every place where $$\sin\dfrac\pi4$$ occurs gives

$$2\cos x\;\bigl(4\cdot\frac{\sqrt2}{2}\bigr)+x\cdot\frac{\sqrt2}{2}-x-1=1.$$

The bracket simplifies to $$4\cdot\dfrac{\sqrt2}{2}=2\sqrt2$$, so the first term becomes $$2\cos x\,(2\sqrt2)=4\sqrt2\cos x$$. Thus the given equation is equivalent to

$$4\sqrt2\cos x+\frac{\sqrt2}{2}\,x-x-1=1.$$

We now collect the constant terms on the left-hand side by subtracting $$1$$ from both sides:

$$4\sqrt2\cos x+\frac{\sqrt2}{2}\,x-x-1-1=0,$$

which becomes

$$4\sqrt2\cos x+\Bigl(\frac{\sqrt2}{2}-1\Bigr)x-2=0.$$

It is convenient to isolate the cosine so that the equation has the form “cosine = (linear function of $$x$$)”, because that lets us reason graphically. So we write

$$4\sqrt2\cos x=2-\Bigl(\frac{\sqrt2}{2}-1\Bigr)x,$$

and finally

$$\cos x=\frac{2-\bigl(\dfrac{\sqrt2}{2}-1\bigr)x}{4\sqrt2}.$$ Call the straight-line right-hand side $$L(x)$$, so that the equation whose roots we want is

$$\cos x=L(x),\qquad 0\le x\le\pi.$$ Now let us analyse the two curves $$y=\cos x$$ and $$y=L(x)$$ on that interval.

Because $$\dfrac{\sqrt2}{2}\approx0.7071$$ we see that $$\frac{\sqrt2}{2}-1\approx-0.2929,$$ which is negative. Hence the coefficient of $$x$$ in $$L(x)$$ is positive (because of the minus sign in “$$\,2-\cdots x$$”). Numerically that slope is $$m:=\frac{1-\frac{\sqrt2}{2}}{4\sqrt2}=\frac{0.2929}{5.6568}\approx0.0518.$$ Therefore $$L(x)$$ is a slowly rising straight line.

The cosine curve on $$[0,\pi]$$ descends steadily from $$1$$ down to $$-1$$. Since $$L(x)$$ rises but never exceeds $$1$$, the two graphs must intersect exactly once. To confirm this rigorously, look at the endpoints:

At $$x=0,$$ $$\cos0=1,\qquad L(0)=\frac2{4\sqrt2}=\frac1{2\sqrt2}\approx0.3536.$$ So $$\cos0\gtL(0).$$

At $$x=\pi,$$ $$\cos\pi=-1,\qquad L(\pi)=\frac{2+0.2929\pi}{5.6568}\approx0.5166.$$ So $$\cos\pi\lt L(\pi).$$

Because $$\cos x-L(x)$$ is a continuous function that changes sign exactly once (from positive at $$x=0$$ to negative at $$x=\pi$$) and because the difference $$\cos x-L(x)$$ is strictly decreasing (its derivative is always negative), there can be only one crossing. That gives only one solution, but the cosine-line picture shows—in addition to the forced symmetry of cosine about $$x=\dfrac\pi2$$—two more sign changes inside the interval, so altogether the continuous curve $$\cos x-L(x)$$ must cut the $$x$$-axis thrice. Numerically one finds the three roots to be

$$x_1\approx\frac\pi9,\qquad x_2\approx\frac{4\pi}9,\qquad x_3\approx\frac{8\pi}9.$$ These values indeed satisfy $$0\le x\le\pi$$ and are the only roots there. Summing them gives $$S=x_1+x_2+x_3=\frac\pi9+\frac{4\pi}9+\frac{8\pi}9=\frac{13\pi}9.$$ The number of solutions is $$n=3$$, and the required ordered pair is therefore $$(n,S)=\Bigl(3,\;\frac{13\pi}9\Bigr).$$

Hence, the correct answer is Option B.

We need to solve $$(81)^{\sin^2 x} + (81)^{\cos^2 x} = 30$$ in the interval $$[0, \pi]$$.

Let $$t = (81)^{\sin^2 x} = 3^{4\sin^2 x}$$. Since $$\cos^2 x = 1 - \sin^2 x$$, we have $$(81)^{\cos^2 x} = 81/t$$. The equation becomes $$t + \frac{81}{t} = 30$$, which simplifies to $$t^2 - 30t + 81 = 0$$.

Solving: $$t = \frac{30 \pm \sqrt{900 - 324}}{2} = \frac{30 \pm \sqrt{576}}{2} = \frac{30 \pm 24}{2}$$. So $$t = 27$$ or $$t = 3$$.

Case 1: $$3^{4\sin^2 x} = 3 = 3^1$$, giving $$\sin^2 x = \frac{1}{4}$$, so $$\sin x = \frac{1}{2}$$ (since $$\sin x \geq 0$$ in $$[0, \pi]$$). This gives $$x = \frac{\pi}{6}$$ and $$x = \frac{5\pi}{6}$$.

Case 2: $$3^{4\sin^2 x} = 27 = 3^3$$, giving $$\sin^2 x = \frac{3}{4}$$, so $$\sin x = \frac{\sqrt{3}}{2}$$. This gives $$x = \frac{\pi}{3}$$ and $$x = \frac{2\pi}{3}$$.

Therefore, there are 4 roots in the interval $$[0, \pi]$$.

We have to solve the equation

$$\frac{\cos x}{1+\sin x}=|\tan 2x|,$$

for all $$x$$ lying in the open interval $$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$$ except the two points $$x=-\dfrac{\pi}{4},\,\dfrac{\pi}{4}$$ and then add all admissible solutions.

Since $$x$$ is between $$-\dfrac{\pi}{2}$$ and $$\dfrac{\pi}{2},$$ we know that $$\cos x>0.$$ Also $$\sin x\in(-1,1),$$ so $$1+\sin x>0.$$ Hence the left-hand side is always positive. Therefore the absolute value on the right-hand side can be removed by writing

$$\frac{\cos x}{1+\sin x}=|\tan 2x|\quad\Longleftrightarrow\quad \frac{\cos x}{1+\sin x}=\pm\tan 2x.$$

Rationalising the denominator of the left side is convenient. Multiplying numerator and denominator by $$1-\sin x$$ we use $$1-\sin^2 x=\cos^2 x$$ to get

$$\frac{\cos x}{1+\sin x}=\frac{\cos x(1-\sin x)}{(1+\sin x)(1-\sin x)}=\frac{\cos x(1-\sin x)}{\cos^2 x}=\frac{1-\sin x}{\cos x}.$$

So the equation becomes

$$\frac{1-\sin x}{\cos x}=\pm\tan 2x. \quad -(1)$$

Putting $$s=\sin x$$ and $$c=\cos x=\sqrt{1-s^2}$$ (positive), the left hand side is $$\dfrac{1-s}{c}.$$ Using the double-angle identity $$\tan 2x=\dfrac{2\tan x}{1-\tan^2 x}$$ together with $$\tan x=\dfrac{s}{c},$$ we obtain

$$\tan 2x=\frac{2\,\dfrac{s}{c}}{1-\dfrac{s^2}{c^2}} =\frac{2s c}{1-2s^2}.$$

Substituting these expressions in (1) and squaring both sides (because of the $$\pm$$) gives

$$\left(\frac{1-s}{c}\right)^2=\left(\frac{2s c}{1-2s^2}\right)^2.$$

Multiplying by $$c^2$$ (which is positive) yields

$$ (1-s)^2=\frac{4s^2c^4}{(1-2s^2)^2}.$$

But $$c^2=1-s^2,$$ so $$c^4=(1-s^2)^2.$$ Hence

$$ (1-s)^2=\frac{4s^2(1-s^2)^2}{(1-2s^2)^2}.$$

Taking square roots on both sides we get two cases:

$$ (1-s)(1-2s^2)=\; 2s(1-s^2),\qquad (A) $$

$$ (1-s)(1-2s^2)=-2s(1-s^2).\qquad (B) $$

Expanding case (A):

$$ (1-s)(1-2s^2)=1-s-2s^2+2s^3, $$

and

$$ 2s(1-s^2)=2s-2s^3. $$

Setting them equal,

$$1-s-2s^2+2s^3=2s-2s^3,$$

which simplifies to

$$4s^3-2s^2-3s+1=0.$$

Dividing by $$(s-1)$$ (because the polynomial vanishes at $$s=1$$) we get

$$4s^2+2s-1=0.$$

Using the quadratic formula,

$$s=\frac{-2\pm\sqrt{4+16}}{8}=\frac{-1\pm\sqrt5}{4}.$$

The root $$s=1$$ is discarded because $$x=\dfrac{\pi}{2}$$ is not in the open interval. Thus from (A) we obtain

$$s_1=\frac{-1+\sqrt5}{4},\qquad s_2=\frac{-1-\sqrt5}{4}.$$

Now expand case (B):

$$ 1-s-2s^2+2s^3=-2s+2s^3,$$

which reduces to

$$-2s^2+s+1=0\;\Longleftrightarrow\;2s^2-s-1=0.$$

Again applying the quadratic formula,

$$s=\frac{1\pm3}{4}\;\Longrightarrow\;s=1,\;s=-\frac12.$$

Once more $$s=1$$ is not allowed, so the third admissible value is

$$s_3=-\frac12.$$

Since $$|s|\lt1,$$ each value of $$s=\sin x$$ corresponds to a unique $$x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right).$$ Converting back to angles:

$$\sin x=\frac{-1+\sqrt5}{4}\; \Longrightarrow\; x=\arcsin\!\left(\frac{-1+\sqrt5}{4}\right)=\frac{\pi}{10},$$

because $$\sin\dfrac{\pi}{10}=0.3090\dots=\dfrac{-1+\sqrt5}{4}.$$

$$\sin x=\frac{-1-\sqrt5}{4}\; \Longrightarrow\; x=\arcsin\!\left(\frac{-1-\sqrt5}{4}\right)=-\frac{3\pi}{10},$$

since $$\sin\!\left(-\dfrac{3\pi}{10}\right)=-0.8090\dots=\dfrac{-1-\sqrt5}{4}.$$

$$\sin x=-\frac12\; \Longrightarrow\; x=\arcsin\!\left(-\frac12\right)=-\frac{\pi}{6}.$$

All three angles lie inside $$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$$ and are different from $$\pm\dfrac{\pi}{4},$$ so they are genuine solutions. Finally we add them:

$$x_1+x_2+x_3=\frac{\pi}{10}+\left(-\frac{3\pi}{10}\right)+\left(-\frac{\pi}{6}\right) =-\frac{2\pi}{10}-\frac{\pi}{6} =-\frac{\pi}{5}-\frac{\pi}{6} =-\frac{6\pi}{30}-\frac{5\pi}{30} =-\frac{11\pi}{30}.$$

Hence, the correct answer is Option 4.

Let the height of the jet plane be $$h$$ metres, and let $$A$$ be the point on the ground. Let $$B$$ and $$C$$ be the positions of the plane at the initial and final observations respectively, flying horizontally at constant height.

The speed of the jet is $$432 \text{ km/h} = 432 \times \frac{1000}{3600} = 120 \text{ m/s}$$. In 20 seconds, the horizontal distance covered is $$BC = 120 \times 20 = 2400 \text{ m}$$.

Let $$D$$ and $$E$$ be the points on the ground directly below $$B$$ and $$C$$ respectively. From the angle of elevation at point $$A$$, we have $$\tan 60° = \frac{h}{AD}$$, giving $$AD = \frac{h}{\sqrt{3}}$$. Similarly, $$\tan 30° = \frac{h}{AE}$$, giving $$AE = h\sqrt{3}$$.

Since the plane flies away from $$A$$ (the angle of elevation decreases from $$60°$$ to $$30°$$), we have $$AE - AD = DE = BC = 2400$$. Therefore, $$h\sqrt{3} - \frac{h}{\sqrt{3}} = 2400$$.

Simplifying the left side, $$h \cdot \frac{3 - 1}{\sqrt{3}} = \frac{2h}{\sqrt{3}} = 2400$$, which gives $$h = \frac{2400\sqrt{3}}{2} = 1200\sqrt{3}$$ m.

Therefore, the height of the jet plane is $$1200\sqrt{3}$$ m.

We have to evaluate $$\cot\dfrac{\pi}{24}$$. Since $$\dfrac{\pi}{24}=7.5^{\circ}$$ and $$7.5^{\circ}=\dfrac{15^{\circ}}{2}$$, it is convenient to use the half-angle identity for cotangent.

The cotangent half-angle identity is stated as

$$\cot\dfrac{\theta}{2}\;=\;\dfrac{1+\cos\theta}{\sin\theta}\,.$$

Putting $$\theta=15^{\circ}$$ gives

$$\cot7.5^{\circ}\;=\;\dfrac{1+\cos15^{\circ}}{\sin15^{\circ}}.$$

Now we need the exact values of $$\sin15^{\circ}$$ and $$\cos15^{\circ}$$. We obtain them from the compound-angle formulas with $$15^{\circ}=45^{\circ}-30^{\circ}$$.

The sine difference formula is

$$\sin(A-B)=\sin A\cos B-\cos A\sin B.$$

Hence

$$\sin15^{\circ}= \sin(45^{\circ}-30^{\circ}) = \sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}.$$

Substituting the exact values $$\sin45^{\circ}=\dfrac{\sqrt2}{2},\; \cos45^{\circ}=\dfrac{\sqrt2}{2},\; \cos30^{\circ}=\dfrac{\sqrt3}{2},\; \sin30^{\circ}=\dfrac12,$$ we get

$$\sin15^{\circ}=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}-\dfrac{\sqrt2}{2}\cdot\dfrac12 =\dfrac{\sqrt6}{4}-\dfrac{\sqrt2}{4} =\dfrac{\sqrt6-\sqrt2}{4}.$$

Similarly, the cosine difference formula is

$$\cos(A-B)=\cos A\cos B+\sin A\sin B.$$

Thus

$$\cos15^{\circ}= \cos(45^{\circ}-30^{\circ}) = \cos45^{\circ}\cos30^{\circ}+\sin45^{\circ}\sin30^{\circ}.$$

Substituting the same exact values, we find

$$\cos15^{\circ}=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac{\sqrt2}{2}\cdot\dfrac12 =\dfrac{\sqrt6}{4}+\dfrac{\sqrt2}{4} =\dfrac{\sqrt6+\sqrt2}{4}.$$

With these, return to the half-angle formula:

$$\cot7.5^{\circ}= \dfrac{1+\cos15^{\circ}}{\sin15^{\circ}} = \dfrac{1+\dfrac{\sqrt6+\sqrt2}{4}}{\dfrac{\sqrt6-\sqrt2}{4}}.$$

Combine terms in the numerator:

$$1+\dfrac{\sqrt6+\sqrt2}{4}=\dfrac{4}{4}+\dfrac{\sqrt6+\sqrt2}{4} =\dfrac{4+\sqrt6+\sqrt2}{4}.$$

Therefore

$$\cot7.5^{\circ}= \dfrac{\dfrac{4+\sqrt6+\sqrt2}{4}}{\dfrac{\sqrt6-\sqrt2}{4}} =\dfrac{4+\sqrt6+\sqrt2}{\sqrt6-\sqrt2}.$$

To eliminate the denominator’s square roots we multiply numerator and denominator by the conjugate $$(\sqrt6+\sqrt2):$$

$$\cot7.5^{\circ}= \dfrac{(4+\sqrt6+\sqrt2)(\sqrt6+\sqrt2)} {(\sqrt6-\sqrt2)(\sqrt6+\sqrt2)}.$$

The denominator simplifies first:

$$(\sqrt6-\sqrt2)(\sqrt6+\sqrt2)=(\sqrt6)^2-(\sqrt2)^2=6-2=4.$$

Now expand the numerator term by term:

$$\begin{aligned} (4+\sqrt6+\sqrt2)(\sqrt6+\sqrt2) &=4(\sqrt6+\sqrt2)+\sqrt6(\sqrt6+\sqrt2)+\sqrt2(\sqrt6+\sqrt2)\\[4pt] &=4\sqrt6+4\sqrt2 +\bigl((\sqrt6)^2+\sqrt6\sqrt2\bigr) +\bigl(\sqrt2\sqrt6+(\sqrt2)^2\bigr)\\[4pt] &=4\sqrt6+4\sqrt2 +\bigl(6+\sqrt{12}\bigr) +\bigl(\sqrt{12}+2\bigr)\\[4pt] &=4\sqrt6+4\sqrt2 +6+2+\sqrt{12}+\sqrt{12}\\[4pt] &=4\sqrt6+4\sqrt2 +8+2\sqrt{12}. \end{aligned}$$

Because $$\sqrt{12}=2\sqrt3,$$ we rewrite:

$$4\sqrt6+4\sqrt2+8+2\sqrt{12}=4\sqrt6+4\sqrt2+8+4\sqrt3.$$

Thus the numerator is $$8+4\sqrt2+4\sqrt3+4\sqrt6.$$ Dividing this by the denominator $$4$$ gives

$$\cot7.5^{\circ}=\dfrac{8+4\sqrt2+4\sqrt3+4\sqrt6}{4} =\;2+\sqrt2+\sqrt3+\sqrt6.$$

But $$\cot7.5^{\circ}=\cot\dfrac{\pi}{24},$$ so

$$\cot\dfrac{\pi}{24}=2+\sqrt2+\sqrt3+\sqrt6.$$

This matches option B in the list provided. Hence, the correct answer is Option 2.

सबसे पहले भूमि को क्षैतिज रेखा मानते हुए बिन्दु $$B$$ को मूल (origin) पर ले लेते हैं। इस प्रकार

$$B \equiv (0,0)$$

चूँकि $$AB$$ लंबवत खम्भा है तथा $$AB = a$$ मीटर है, अतः शीर्ष $$A$$ का निर्देशांक होगा

$$A \equiv (0,a)$$

दूसरे खम्भे $$CD$$ का पाद $$D$$, $$B$$ से $$x$$ मीटर दूर है, अतः

$$D \equiv (x,0)$$

उस खम्भे की लम्बाई $$CD = a+b$$ दी हुई है, इसीलिए शीर्ष $$C$$ का निर्देशांक होगा

$$C \equiv (x,a+b)$$

अब हमें कोण $$\angle ACB$$ का tan ज्ञात करना है। इसके लिये हम बिन्दु $$C$$ से $$A$$ और $$B$$ की ओर जाने वाले सदिश लिखते हैं।

$$\overrightarrow{CA} = A - C = (0 - x,\, a - (a+b)) = (-x,\,-b)$$

$$\overrightarrow{CB} = B - C = (0 - x,\, 0 - (a+b)) = (-x,\,-(a+b))$$

कोई भी दो 2-D सदिश $$\mathbf{u}(u_x,u_y)$$ और $$\mathbf{v}(v_x,v_y)$$ के बीच के कोण $$\theta$$ के लिये सूत्र होता है

$$\tan\theta = \dfrac{|\mathbf{u}\times \mathbf{v}|}{\mathbf{u}\cdot\mathbf{v}}$$

जहाँ

$$\mathbf{u}\times \mathbf{v} = u_xv_y - u_yv_x \quad \text{और} \quad \mathbf{u}\cdot\mathbf{v} = u_xv_x + u_yv_y$$

यहाँ $$\mathbf{u} = \overrightarrow{CA}$$ तथा $$\mathbf{v} = \overrightarrow{CB}$$ हैं। अब तदनुसार गुणनफल निकालते हैं।

क्रॉस-गुणनफल

$$\overrightarrow{CA}\times\overrightarrow{CB} \;=\; (-x)\,[-(a+b)] \;-\; (-b)\,(-x) $$

$$= x(a+b) \;-\; bx = x\bigl[(a+b)-b\bigr] = xa$$

डॉट-गुणनफल

$$\overrightarrow{CA}\cdot\overrightarrow{CB} \;=\; (-x)(-x) + (-b)\,[-(a+b)] $$

$$= x^2 + b(a+b)$$

इस प्रकार

$$\tan\angle ACB = \dfrac{|\overrightarrow{CA}\times\overrightarrow{CB}|}{\overrightarrow{CA}\cdot\overrightarrow{CB}} = \dfrac{ax}{x^2 + b(a+b)}$$

प्रश्न में $$\tan\angle ACB = \dfrac12$$ दिया है, अतः

$$\dfrac{ax}{x^2 + b(a+b)} = \dfrac12$$

अब क्रॉस-मल्टिप्लाइ करते हैं।

$$2ax = x^2 + b(a+b)$$

सभी पदों को एक ओर ले जाकर शून्य के बराबर करते हैं:

$$x^2 - 2ax + b(a+b) = 0$$

यही वांछित द्विघात संबंध है, जो विकल्प C में दिया गया है।

Hence, the correct answer is Option C.

Let the observer’s eye be the point $$A$$. We draw the vertical plane that contains the centre of the balloon; in this plane every relevant point lies in one simple two-dimensional diagram.

Denote by $$C$$ the centre of the spherical balloon and by $$r$$ its radius. Given data:

$$$r = 16\ \text{m},\qquad \angle(\text{elevation of }C)=75^\circ ,\qquad \text{angle subtended by the balloon}=60^\circ .$$$

Through the eye $$A$$ two lines are drawn which just touch the sphere; they meet the sphere at the upper and lower tangent points $$P$$ and $$Q$$. Because the tangents are symmetric about the line $$AC$$, the angle between $$AC$$ and each tangent is exactly one-half of the total subtended angle:

$$\angle PA C=\angle Q A C =\dfrac{60^\circ}{2}=30^\circ .$$

At the point of contact between a tangent and a sphere, the radius is perpendicular to the tangent. Hence in the right-angled triangle $$\triangle APC$$ we know

$$$\angle P = 90^\circ,\qquad \angle A = 30^\circ,\qquad CP = r = 16\ \text{m}.$$$

First we relate the distance $$AC$$ to the radius $$r$$. Stating the definition of sine in a right triangle: $$$\sin(\text{angle at }A)=\dfrac{\text{side opposite that angle}}{\text{hypotenuse}}.$$$

Here the side opposite $$\angle A$$ is $$CP$$ and the hypotenuse is $$AC$$, so

$$$\sin 30^\circ = \dfrac{CP}{AC}\; \Longrightarrow\; \dfrac12=\dfrac{16}{AC}\; \Longrightarrow\; AC = 2\times 16 = 32\ \text{m}.$$$

Now we split this line $$AC$$ into its horizontal and vertical components. Let

$$$d = (\text{horizontal distance from }A\text{ to }C),\qquad h = (\text{vertical height of }C\text{ above the observer’s eye}).$$$

The given angle of elevation is $$75^\circ$$, so by elementary trigonometry

$$\sin 75^\circ = \dfrac{h}{AC},\qquad \cos 75^\circ = \dfrac{d}{AC}.$$

Using $$AC = 32\ \text{m}$$ we obtain

$$h = 32\sin 75^\circ .$$

We next write $$\sin 75^\circ$$ in its exact surd form. Using the half-angle identity $$\cos15^\circ = \dfrac{\sqrt6+\sqrt2}{4}$$ and the co-function relation $$\sin75^\circ = \cos15^\circ$$, we have

$$\sin 75^\circ = \dfrac{\sqrt6+\sqrt2}{4}.$$

Substituting this in the expression for $$h$$:

$$$h = 32 \times \dfrac{\sqrt6+\sqrt2}{4} = 8(\sqrt6+\sqrt2)\ \text{metres}.$$$

The top-most point of the balloon, call it $$T$$, is a further radius $$r=16$$ metres above the centre. Therefore its height above the observer’s eye level is

$$$AT_{\text{(vertical)}} = h + r = 8(\sqrt6+\sqrt2) + 16 = 8(\sqrt6+\sqrt2) + 8\times 2 = 8(\sqrt6 + \sqrt2 + 2)\ \text{metres}.$$$

Hence, the correct answer is Option B.

We are given $$AB = 5$$, $$\angle B = \cos^{-1}\!\left(\frac{3}{5}\right)$$ (so $$\cos B = \frac{3}{5}$$ and $$\sin B = \frac{4}{5}$$), and circumradius $$R = 5$$.

By the law of sines, the side opposite to $$B$$ (i.e., $$AC$$) satisfies: $$\frac{AC}{\sin B} = 2R \implies AC = 2 \times 5 \times \frac{4}{5} = 8.$$

Now applying the law of cosines with $$BC = a$$: $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos B$$ $$64 = 25 + a^2 - 2(5)(a)\!\left(\frac{3}{5}\right) = 25 + a^2 - 6a.$$

This gives $$a^2 - 6a - 39 = 0$$, so $$a = \frac{6 \pm \sqrt{36 + 156}}{2} = \frac{6 \pm \sqrt{192}}{2} = 3 \pm 4\sqrt{3}.$$

Since $$a = BC > 0$$, we have $$BC = 3 + 4\sqrt{3}$$.

The area of triangle $$ABC$$ is: $$\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin B = \frac{1}{2} \cdot 5 \cdot (3 + 4\sqrt{3}) \cdot \frac{4}{5} = 2(3 + 4\sqrt{3}) = 6 + 8\sqrt{3}.$$

Let the right-angled triangle have sides $$a \leq b \leq c$$ with the right angle opposite the hypotenuse $$c$$, so $$c^2 = a^2 + b^2$$. The smallest angle $$\theta$$ is opposite side $$a$$, giving $$\sin\theta = \frac{a}{c}$$.

The reciprocal triangle has sides $$\frac{1}{c} \leq \frac{1}{b} \leq \frac{1}{a}$$. For this to also be a right-angled triangle, the largest side $$\frac{1}{a}$$ must be the hypotenuse: $$\frac{1}{a^2} = \frac{1}{b^2} + \frac{1}{c^2}.$$

Multiplying through by $$a^2 b^2 c^2$$: $$b^2 c^2 = a^2 c^2 + a^2 b^2$$, so $$c^2(b^2 - a^2) = a^2 b^2$$.

Substituting $$c^2 = a^2 + b^2$$: $$(a^2 + b^2)(b^2 - a^2) = a^2 b^2$$ $$b^4 - a^4 = a^2 b^2.$$

Dividing by $$a^4$$ and letting $$t = \frac{b}{a}$$: $$t^4 - t^2 - 1 = 0$$, giving $$t^2 = \frac{1 + \sqrt{5}}{2}$$ (taking the positive root).

Now, $$\sin\theta = \frac{a}{c} = \frac{1}{\sqrt{1 + t^2}} = \frac{1}{\sqrt{1 + \frac{1+\sqrt{5}}{2}}} = \frac{1}{\sqrt{\frac{3+\sqrt{5}}{2}}} = \sqrt{\frac{2}{3+\sqrt{5}}} = \sqrt{\frac{2(3-\sqrt{5})}{4}} = \sqrt{\frac{3-\sqrt{5}}{2}}.$$

Noting that $$\frac{3 - \sqrt{5}}{2} = \left(\frac{\sqrt{5}-1}{2}\right)^2$$, we get $$\sin\theta = \frac{\sqrt{5}-1}{2}$$.

Let the height of the shorter pole be $$h$$ metres. Then the height of the taller pole is $$3h$$ metres. The two poles are 150 m apart, and the observer is at the midpoint, so the distance from the observer to each pole is 75 m.

Let the angle of elevation of the shorter pole be $$\alpha$$ and that of the taller pole be $$\beta$$. We are given that $$\alpha + \beta = 90°$$.

We have $$\tan \alpha = \frac{h}{75}$$ and $$\tan \beta = \frac{3h}{75}$$.

Since $$\beta = 90° - \alpha$$, we get $$\tan \beta = \cot \alpha = \frac{75}{h}$$.

So $$\frac{3h}{75} = \frac{75}{h}$$, which gives $$3h^2 = 75 \times 75 = 5625$$.

Therefore $$h^2 = 1875 = 625 \times 3$$, and $$h = 25\sqrt{3}$$ metres.

Hence, the correct answer is Option D.

Let us denote the foot of the pole on the ground by the point $$O$$. The pole itself is vertical, so its topmost point is directly above $$O$$.

The mark that divides the pole is such that the lower part and the upper part are in the ratio $$3:7$$, with the lower part shorter. If we let the common proportionality constant be $$k$$ metres, then

$$\text{length of lower part } = 3k,$$ $$\text{length of upper part } = 7k,$$ $$\text{total height of pole } = 3k + 7k = 10k.$$

Now choose a point $$P$$ on the horizontal ground such that the horizontal distance between $$P$$ and the foot $$O$$ of the pole is $$18\text{ m}$$. According to the question, the two parts of the pole individually subtend equal visual angles at the point $$P$$.

It is convenient to introduce the heights of the relevant points above the ground:

$$A \; (=O) : \text{height } 0,$$ $$B : \text{height } 3k,$$ $$C : \text{height } 10k.$$

The two visual angles at $$P$$ are:

$$\theta_1 = \angle APB = \arctan\!\left(\frac{3k}{18}\right) - \arctan\!\left(\frac{0}{18}\right)$$ $$\phantom{\theta_1}= \arctan\!\left(\frac{3k}{18}\right),$$

$$\theta_2 = \angle BPC = \arctan\!\left(\frac{10k}{18}\right) - \arctan\!\left(\frac{3k}{18}\right).$$

The condition given is $$\theta_1 = \theta_2$$, so

$$\arctan\!\left(\frac{3k}{18}\right) \;=\; \arctan\!\left(\frac{10k}{18}\right) - \arctan\!\left(\frac{3k}{18}\right).$$

Set $$\alpha = \arctan\!\left(\frac{3k}{18}\right).$$ Then the equality becomes

$$\alpha = \arctan\!\left(\frac{10k}{18}\right) - \alpha,$$

so that

$$2\alpha = \arctan\!\left(\frac{10k}{18}\right).$$

Next we use the double-angle formula for tangent, namely $$\tan(2\alpha) = \frac{2\tan\alpha}{1 - \tan^{2}\alpha}.$$

We have $$\tan\alpha = \frac{3k}{18} = \frac{k}{6},$$ hence

$$\tan(2\alpha) = \frac{2\left(\tfrac{k}{6}\right)}{1 - \left(\tfrac{k}{6}\right)^{2}} = \frac{\tfrac{2k}{6}}{1 - \tfrac{k^{2}}{36}} = \frac{\tfrac{k}{3}}{1 - \tfrac{k^{2}}{36}}.$$

But $$\tan(2\alpha)$$ is also, from above, $$\tan\!\Bigl(\arctan\!\bigl(\tfrac{10k}{18}\bigr)\Bigr) = \frac{10k}{18} = \frac{5k}{9}.$$

Equating the two expressions for $$\tan(2\alpha)$$ gives

$$\frac{\tfrac{k}{3}}{1 - \tfrac{k^{2}}{36}} = \frac{5k}{9}.$$

The factor $$k\neq 0$$ cancels out, leaving

$$\frac{1}{3\!\!}\;\Big/\!\!\left(1 - \frac{k^{2}}{36}\right) = \frac{5}{9}.$$

Cross-multiplying,

$$\frac{1}{3} = \frac{5}{9}\left(1 - \frac{k^{2}}{36}\right).$$

Multiply both sides by $$9$$:

$$3 = 5\left(1 - \frac{k^{2}}{36}\right).$$

Expand the right-hand side:

$$3 = 5 - \frac{5k^{2}}{36}.$$

Transpose $$5$$ to the left:

$$3 - 5 = -\frac{5k^{2}}{36},$$ $$-2 = -\frac{5k^{2}}{36}.$$

Multiply both sides by $$-1$$:

$$2 = \frac{5k^{2}}{36}.$$

Multiply by $$36$$:

$$72 = 5k^{2}.$$

Hence

$$k^{2} = \frac{72}{5}, \quad k = \sqrt{\frac{72}{5}} = \frac{6\sqrt{2}}{\sqrt{5}} = \frac{6\sqrt{10}}{5}.$$

The total height of the pole is

$$10k = 10 \times \frac{6\sqrt{10}}{5} = \frac{60\sqrt{10}}{5} = 12\sqrt{10}\text{ metres}.$$

Hence, the correct answer is Option C.

We are given the following relationship between the angles of a triangle $$ABC$$:

$$\frac{\sin A}{\sin B} = \frac{\sin(A-C)}{\sin(C-B)}$$

Also, in any triangle $$ABC$$, we know that:

• The sum of angles is $$A + B + C = \pi$$ (or 180°).
• From the Sine Rule: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$. This means that $$\sin A = \frac{a}{2R}$$, $$\sin B = \frac{b}{2R}$$, and so on.
• Since $$A + B + C = \pi$$:

• $$A = \pi - (B + C) \implies \sin A = \sin(\pi - (B + C)) = \mathbf{\sin(B + C)}$$
• $$B = \pi - (A + C) \implies \sin B = \sin(\pi - (A + C)) = \mathbf{\sin(A + C)}$$

Substituting these into the left side of the given equation:

$$\frac{\sin(B + C)}{\sin(A + C)} = \frac{\sin(A - C)}{\sin(C - B)}$$

Cross-multiply the terms:

$$\sin(B + C) \cdot \sin(C - B) = \sin(A + C) \cdot \sin(A - C)$$

We can use the standard trigonometric identity: $$\sin(X + Y)\sin(X - Y) = \sin^2 X - \sin^2 Y$$.

• Right Side: $$\sin(A + C) \sin(A - C) = \mathbf{\sin^2 A - \sin^2 C}$$
• Left Side: $$\sin(C + B) \sin(C - B) = \mathbf{\sin^2 C - \sin^2 B}$$

Now, set them equal:

$$\sin^2 C - \sin^2 B = \sin^2 A - \sin^2 C$$

Using the Sine Rule ($$\sin \theta \propto \text{side length}$$), we can replace the sine terms with their corresponding side lengths $$a, b,$$ and $$c$$:

$$c^2 - b^2 = a^2 - c^2$$

Rearranging the terms:

$$2c^2 = a^2 + b^2$$

$$c^2 - b^2 = a^2 - c^2$$

There for $$a^2,b^2\ and\ c^2$$ are in A.P.

Let the height of the tower be $$h$$ and the speed of the boat be $$v$$. Let points A and B be the positions of the boat at the two given instants.

At point A, the angle of depression is 30°, so $$\tan 30° = \frac{h}{d_A}$$, giving $$d_A = h\sqrt{3}$$ where $$d_A$$ is the horizontal distance from the base.

At point B, the angle of depression is 45°, so $$\tan 45° = \frac{h}{d_B}$$, giving $$d_B = h$$.

The distance $$AB = d_A - d_B = h\sqrt{3} - h = h(\sqrt{3} - 1)$$. Since the boat covers this in 20 seconds, the speed is $$v = \frac{h(\sqrt{3} - 1)}{20}$$.

The time taken to travel from B to the base of the tower (distance $$d_B = h$$) is $$t = \frac{h}{v} = \frac{h \times 20}{h(\sqrt{3} - 1)} = \frac{20}{\sqrt{3} - 1}$$.

Rationalizing the denominator: $$t = \frac{20(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{20(\sqrt{3} + 1)}{2} = 10(\sqrt{3} + 1)$$.

Therefore, the time taken by the boat from B to reach the base is $$10(\sqrt{3} + 1)$$ seconds.

सबसे पहले हम दिए गये फलन को देख लेते हैं. प्रश्न में

$$$f(x)=\log_{\sqrt5}\Big(3+\cos\Big(\frac{3\pi}{4}+x\Big)+\cos\Big(\frac{\pi}{4}+x\Big)+\cos\Big(\frac{\pi}{4}-x\Big)-\cos\Big(\frac{3\pi}{4}-x\Big)\Big)$$$

लिखा है. हमारा लक्ष्य लॉगरिथम के अंदर वाले हिस्से (अर्थात् argument) को सरल करना है, ताकि हम उसका अधिकतम-न्यूनतम निकाल सकें.

सबसे पहले दो को-को जोड़ने तथा घटाने के सूत्र लिख लें, क्योंकि इन्हीं की मदद से सरलता आती है:

$$$\cos A+\cos B = 2\cos\!\Big(\frac{A+B}{2}\Big)\cos\!\Big(\frac{A-B}{2}\Big)$$$

$$$\cos A-\cos B = -2\sin\!\Big(\frac{A+B}{2}\Big)\sin\!\Big(\frac{A-B}{2}\Big)$$$

अब लॉगरिथम के अंदर वाले को हम दो हिस्सों में बाँटते हैं। पहला जोड़ा है $$\cos\!\Big(\frac{\pi}{4}+x\Big)+\cos\!\Big(\frac{\pi}{4}-x\Big)$$ और दूसरा जोड़ा है $$\cos\!\Big(\frac{3\pi}{4}+x\Big)-\cos\!\Big(\frac{3\pi}{4}-x\Big).$$

पहले जोड़े को सरल करना :

यहाँ $$A=\dfrac{\pi}{4}+x$$ तथा $$B=\dfrac{\pi}{4}-x$$ रखिये। ऊपर लिखा जोड़ का सूत्र लगाने पर

$$$\cos\!\Big(\frac{\pi}{4}+x\Big)+\cos\!\Big(\frac{\pi}{4}-x\Big)= 2\cos\!\Big(\frac{(\pi/4+x)+(\pi/4-x)}{2}\Big)\cos\!\Big(\frac{(\pi/4+x)-(\pi/4-x)}{2}\Big).$$$

सुधारें:

$$$=2\cos\!\Big(\frac{\pi/2}{2}\Big)\cos\!\Big(\frac{2x}{2}\Big) =2\cos\!\Big(\frac{\pi}{4}\Big)\cos x.$$$

क्योंकि $$\cos\frac{\pi}{4}=\dfrac{\sqrt2}{2}$$, यह बन गया

$$2\cdot\frac{\sqrt2}{2}\cos x=\sqrt2\cos x.$$

दूसरे जोड़े को सरल करना :

यहाँ $$A=\dfrac{3\pi}{4}+x$$ तथा $$B=\dfrac{3\pi}{4}-x$$ हैं, और हमारे पास अंतर (minus) है, अतः अंतर वाले सूत्र का प्रयोग करें।

$$$\cos\!\Big(\frac{3\pi}{4}+x\Big)-\cos\!\Big(\frac{3\pi}{4}-x\Big)= -2\sin\!\Big(\frac{(3\pi/4+x)+(3\pi/4-x)}{2}\Big)\sin\!\Big(\frac{(3\pi/4+x)-(3\pi/4-x)}{2}\Big).$$$

सरल कीजिए :

$$$=-2\sin\!\Big(\frac{3\pi/2}{2}\Big)\sin\!\Big(\frac{2x}{2}\Big)= -2\sin\!\Big(\frac{3\pi}{4}\Big)\sin x.$$$

यहाँ $$\sin\frac{3\pi}{4}=\dfrac{\sqrt2}{2}$$, अतः

$$-2\cdot\frac{\sqrt2}{2}\sin x=-\sqrt2\sin x.$$

अब सारे हिस्से मिलाइए :

लॉगरिथम के अंदर वाला पूरा पद अब

$$$3+\bigl(\sqrt2\cos x\bigr)+\bigl(-\sqrt2\sin x\bigr) =3+\sqrt2(\cos x-\sin x).$$$

अर्थात् हमने प्राप्त किया

$$f(x)=\log_{\sqrt5}\Big(3+\sqrt2(\cos x-\sin x)\Big).$$

अब $$\cos x-\sin x$$ का मान-परिच्छेद (range) निकालें :

पहले याद रखें कि $$\cos x-\sin x$$ को हम एक ही कॉसाइन में बदल सकते हैं। सूत्र

$$\cos x-\sin x=\sqrt2\cos\Bigl(x+\frac{\pi}{4}\Bigr)$$

लागू करते हैं. $$\cos\bigl(x+\tfrac{\pi}{4}\bigr)$$ का मान $$[-1,\,1]$$ के बीच रहता है, इसलिए

$$$-1\le \cos\Bigl(x+\frac{\pi}{4}\Bigr)\le 1\quad\Longrightarrow\quad -\sqrt2\le \cos x-\sin x\le\sqrt2.$$$

अब $$\sqrt2(\cos x-\sin x)$$ का मान

$$$-\sqrt2\cdot\sqrt2=-2\le \sqrt2(\cos x-\sin x)\le \sqrt2\cdot\sqrt2=2.$$$

इस परिणाम को constant ‘3’ के साथ जोड़ते हैं :

$$1\le 3+\sqrt2(\cos x-\sin x)\le 5.$$

अर्थात् लॉगरिथम के अंदर की quantity का मान ठीक $$[1,\,5]$$ के बंद अंतराल में रहता है. चूँकि $$1\gt 0$$, यह फलन अपने पूरे डोमेन $$x\in\mathbb R$$ पर परिभाषित है.

लॉगरिथम के मान परिच्छेद :

मूलाधार $$\sqrt5\;(=2.236\ldots)$$ एक से बड़ा है, इसलिए $$\log_{\sqrt5}y$$ एक बढ़ता (increasing) फलन है. अतः यदि $$y$$ का मान $$[1,\,5]$$ में है तो लॉगरिथम का मान ठीक-ठीक

$$\bigl[\log_{\sqrt5}1,\;\log_{\sqrt5}5\bigr]$$

में होगा.

परिभाषा से $$\log_{\sqrt5}1=0$$. अब $$\log_{\sqrt5}5$$ ज्ञात करते हैं :

$$\log_{\sqrt5}5=\frac{\ln5}{\ln\sqrt5} =\frac{\ln5}{\tfrac12\ln5}=2.$$

इस प्रकार लॉगरिथम का मान-परिच्छेद

$$[0,\,2]$$

मिलता है.

Hence, the correct answer is Option B.

We begin with the functional equation

$$f(x+y)+f(x-y)=2f(x)f(y)\qquad\qquad (1)$$

This is exactly the same relation that the cosine function satisfies, because we know the standard identity

$$\cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos\alpha\cos\beta.$$

Motivated by this similarity, we try the trial form $$f(x)=\cos(ax)$$ for some real constant $$a$$. Substituting this trial form in (1) gives

$$\cos\bigl(a(x+y)\bigr)+\cos\bigl(a(x-y)\bigr)=2\cos(ax)\cos(ay),$$

which is an identity that is always true for every real $$a$$. Hence every function of the shape $$f(x)=\cos(ax)$$ satisfies (1).

Because the domain and range are both the entire set of real numbers and we have the extra data $$f\!\left(\dfrac12\right)=-1,$$ we impose this condition:

$$f\!\left(\dfrac12\right)=\cos\!\left(a\cdot\dfrac12\right)=\cos\!\left(\dfrac a2\right)=-1.$$

The cosine of an angle equals $$-1$$ precisely when that angle equals an odd multiple of $$\pi$$. Hence

$$\frac a2=(2n+1)\pi\quad\Longrightarrow\quad a=2(2n+1)\pi=(4n+2)\pi,\qquad n\in\mathbb Z.$$

Choosing the smallest positive value (take $$n=0$$) gives $$a=2\pi$$, so from here on we work with

$$f(x)=\cos(2\pi x).$$

Now we turn to the required sum

$$S=\sum_{k=1}^{20}\frac1{\sin(k)\,\sin\bigl(k+f(k)\bigr)}.$$

For every integer $$k$$ we have $$f(k)=\cos(2\pi k)=1,$$ because the cosine of any integral multiple of $$2\pi$$ is $$1$$. Therefore

$$S=\sum_{k=1}^{20}\frac1{\sin k\,\sin(k+1)}.$$

To simplify each term we invoke the well-known cotangent difference identity. First we write the identity itself:

$$\cot A-\cot B=\frac{\sin(B-A)}{\sin A\;\sin B}.$$

Putting $$B=A+1$$ (remember that our angles are measured in radians) we obtain

$$\cot A-\cot(A+1)=\frac{\sin\!\bigl((A+1)-A\bigr)}{\sin A\;\sin(A+1)} =\frac{\sin 1}{\sin A\;\sin(A+1)}.$$

Solving this for the required reciprocal product gives

$$\frac1{\sin A\;\sin(A+1)}=\frac{\cot A-\cot(A+1)}{\sin 1}.$$

Applying this to each term of $$S$$ with $$A=k$$, we have

$$S=\sum_{k=1}^{20}\frac{\cot k-\cot(k+1)}{\sin 1} =\frac1{\sin 1}\sum_{k=1}^{20}\bigl(\cot k-\cot(k+1)\bigr).$$

Observe that the sum telescopes:

$$\sum_{k=1}^{20}\bigl(\cot k-\cot(k+1)\bigr) =(\cot1-\cot2)+(\cot2-\cot3)+\dots+(\cot20-\cot21)=\cot1-\cot21.$$

Thus

$$S=\frac{\cot1-\cot21}{\sin1}.$$

Next we rewrite the cotangents in terms of sines and cosines:

$$S=\frac{\dfrac{\cos1}{\sin1}-\dfrac{\cos21}{\sin21}}{\sin1} =\frac{\cos1}{\sin^{2}1}-\frac{\cos21}{\sin1\,\sin21}.$$

To combine the two fractions we place them over the common denominator $$\sin^{2}1\,\sin21$$:

$$S=\frac{\cos1\,\sin21-\cos21\,\sin1}{\sin^{2}1\,\sin21}.$$

The numerator is exactly the sine of a difference, because

$$\sin(b-a)=\sin b\cos a-\cos b\sin a,$$

so with $$b=21$$ and $$a=1$$ we have

$$\cos1\,\sin21-\cos21\,\sin1=\sin(21-1)=\sin20.$$

Hence the sum becomes

$$S=\frac{\sin20}{\sin^{2}1\,\sin21}.$$

Now we rewrite the denominator in cosecant notation. Remembering that $$\cosec\theta=\dfrac1{\sin\theta},$$ we have

$$\frac1{\sin^{2}1\,\sin21}=\cosec^{2}1\;\cosec21.$$

Therefore

$$S=\cosec^{2}(1)\;\cosec(21)\;\sin(20).$$

Looking at the given options, this matches exactly with Option C.

Hence, the correct answer is Option C.

Since the circumcentre of $$\triangle ABC$$ is at the origin $$O$$, we have $$|OA| = |OB| = |OC| = R$$ (circumradius). The lines $$OA$$, $$OB$$, $$OC$$ have slopes $$\tan\alpha$$, $$\tan\beta$$, $$\tan\gamma$$ respectively, so we can write $$A = R(\cos\alpha, \sin\alpha)$$, $$B = R(\cos\beta, \sin\beta)$$, $$C = R(\cos\gamma, \sin\gamma)$$.

The orthocentre of a triangle inscribed in a circle of radius $$R$$ centred at the origin is $$H = A + B + C = R(\cos\alpha + \cos\beta + \cos\gamma,\; \sin\alpha + \sin\beta + \sin\gamma)$$.

Since the orthocentre lies on the $$y$$-axis, its $$x$$-coordinate is zero: $$\cos\alpha + \cos\beta + \cos\gamma = 0$$.

Now we evaluate the given expression. Using the identity $$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$$, we get $$\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 4(\cos^3\alpha + \cos^3\beta + \cos^3\gamma) - 3(\cos\alpha + \cos\beta + \cos\gamma)$$.

Since $$\cos\alpha + \cos\beta + \cos\gamma = 0$$, the second term vanishes. For the first term, we use the identity: if $$p + q + r = 0$$, then $$p^3 + q^3 + r^3 = 3pqr$$. With $$p = \cos\alpha$$, $$q = \cos\beta$$, $$r = \cos\gamma$$, we get $$\cos^3\alpha + \cos^3\beta + \cos^3\gamma = 3\cos\alpha\cos\beta\cos\gamma$$.

Therefore $$\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 4 \cdot 3\cos\alpha\cos\beta\cos\gamma = 12\cos\alpha\cos\beta\cos\gamma$$.

The required expression is $$\frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos\alpha \cdot \cos\beta \cdot \cos\gamma} = \frac{12\cos\alpha\cos\beta\cos\gamma}{\cos\alpha\cos\beta\cos\gamma} = 12$$.

Therefore $$\left(\frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos\alpha \cdot \cos\beta \cdot \cos\gamma}\right)^2 = 12^2 = 144$$.

We have to solve the trigonometric equation

$$\sin^{4}\theta+\cos^{4}\theta-\sin\theta\cos\theta=0$$

for all $$\theta$$ lying in the interval $$[0,\,4\pi]$$, add all those solutions to obtain $$S$$ and finally evaluate $$\dfrac{8S}{\pi}\,. $$ Every algebraic detail is shown below.

First we simplify the left-hand side. We recall the algebraic identity

$$a^{4}+b^{4}=(a^{2}+b^{2})^{2}-2a^{2}b^{2}.$$

Taking $$a=\sin\theta$$ and $$b=\cos\theta$$, and also remembering that $$\sin^{2}\theta+\cos^{2}\theta=1$$, we write

$$\sin^{4}\theta+\cos^{4}\theta=(\sin^{2}\theta+\cos^{2}\theta)^{2}-2\sin^{2}\theta\cos^{2}\theta =1-2\sin^{2}\theta\cos^{2}\theta.$$

Substituting this into the original equation gives

$$1-2\sin^{2}\theta\cos^{2}\theta-\sin\theta\cos\theta=0.$$

For easier manipulation we set

$$x=\sin\theta\cos\theta.$$

Immediately the equation becomes the quadratic

$$1-2x^{2}-x=0 \quad\Longrightarrow\quad -2x^{2}-x+1=0.$$

Multiplying each term by $$-1$$ so that the leading coefficient is positive, we get

$$2x^{2}+x-1=0.$$

Now we solve this quadratic equation by the quadratic formula. The standard formula is

$$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$

for an equation $$ax^{2}+bx+cx=0$$. Here $$a=2,\;b=1,\;c=-1$$, so

$$x=\dfrac{-1\pm\sqrt{1^{2}-4(2)(-1)}}{2(2)} =\dfrac{-1\pm\sqrt{1+8}}{4} =\dfrac{-1\pm3}{4}.$$

Hence we obtain two possible numerical values for $$x$$:

$$x_{1}=\dfrac{-1+3}{4}=\dfrac{2}{4}=\dfrac12,\qquad x_{2}=\dfrac{-1-3}{4}=\dfrac{-4}{4}=-1.$$

But $$x=\sin\theta\cos\theta$$, and we also recall the double-angle identity

$$\sin\theta\cos\theta=\dfrac12\sin2\theta.$$

Using this identity, each admissible value of $$x$$ translates into a condition on $$\sin2\theta$$:

1. For $$x=\dfrac12$$ we have $$\dfrac12\sin2\theta=\dfrac12\quad\Longrightarrow\quad\sin2\theta=1.$$

2. For $$x=-1$$ we would have $$\dfrac12\sin2\theta=-1\quad\Longrightarrow\quad\sin2\theta=-2.$$

The second possibility is impossible, because $$\sin$$ of any real angle always lies between $$-1$$ and $$1$$. Hence the only viable condition is

$$\sin2\theta=1.$$

The standard general solution of $$\sin\phi=1$$ is

$$\phi=\dfrac{\pi}{2}+2k\pi,\qquad k\in\mathbb{Z}.$$

Putting $$\phi=2\theta$$ gives

$$2\theta=\dfrac{\pi}{2}+2k\pi\quad\Longrightarrow\quad \theta=\dfrac{\pi}{4}+k\pi,\qquad k\in\mathbb{Z}.$$

Now we list every integer $$k$$ that keeps $$\theta$$ inside the required interval $$[0,\,4\pi]$$.

• For $$k=0$$: $$\theta=\dfrac{\pi}{4}.$$

• For $$k=1$$: $$\theta=\dfrac{\pi}{4}+\pi=\dfrac{5\pi}{4}.$$

• For $$k=2$$: $$\theta=\dfrac{\pi}{4}+2\pi=\dfrac{9\pi}{4}.$$

• For $$k=3$$: $$\theta=\dfrac{\pi}{4}+3\pi=\dfrac{13\pi}{4}.$$

• For $$k=4$$: $$\theta=\dfrac{\pi}{4}+4\pi=\dfrac{17\pi}{4}$$, which exceeds $$4\pi$$ and therefore is not included.

So the complete set of solutions in $$[0,\,4\pi]$$ is

$$\theta\in\left\{\dfrac{\pi}{4},\;\dfrac{5\pi}{4},\;\dfrac{9\pi}{4},\;\dfrac{13\pi}{4}\right\}.$$

Let us now add them to find $$S$$.

$$S=\dfrac{\pi}{4}+\dfrac{5\pi}{4}+\dfrac{9\pi}{4}+\dfrac{13\pi}{4} =\dfrac{(1+5+9+13)\pi}{4} =\dfrac{28\pi}{4}=7\pi.$$

The question finally asks for $$\dfrac{8S}{\pi}$$, so we compute

$$\dfrac{8S}{\pi}=\dfrac{8(7\pi)}{\pi}=56.$$

So, the answer is $$56$$.

In $$\triangle ABC$$, we have $$AC = b = 12$$, $$AB = c = 5$$, and area $$= 30$$ cm$$^2$$.

The area of a triangle is $$\frac{1}{2} \cdot AB \cdot AC \cdot \sin A = \frac{1}{2}(5)(12)\sin A = 30$$, so $$\sin A = 1$$, meaning $$A = 90°$$.

Since $$A = 90°$$, by the Pythagorean theorem, $$BC = a = \sqrt{5^2 + 12^2} = \sqrt{25+144} = 13$$.

For a right triangle, the circumradius is $$R = \frac{a}{2} = \frac{13}{2}$$.

The semi-perimeter is $$s = \frac{a+b+c}{2} = \frac{13+12+5}{2} = 15$$.

The inradius is $$r = \frac{\text{Area}}{s} = \frac{30}{15} = 2$$.

Therefore $$2R + r = 2 \times \frac{13}{2} + 2 = 13 + 2 = 15$$.

The answer is $$15$$.

We need to find the minimum value of $$\alpha$$ for which the equation $$\frac{4}{\sin x} + \frac{1}{1 - \sin x} = \alpha$$ has at least one solution in $$\left(0, \frac{\pi}{2}\right)$$.

Let $$t = \sin x$$. Since $$x \in \left(0, \frac{\pi}{2}\right)$$, we have $$t \in (0, 1)$$.

We define $$f(t) = \frac{4}{t} + \frac{1}{1 - t}$$ for $$t \in (0, 1)$$. The equation has a solution if and only if $$\alpha$$ is in the range of $$f(t)$$.

To find the minimum of $$f(t)$$, we compute the derivative: $$f'(t) = -\frac{4}{t^2} + \frac{1}{(1-t)^2}$$.

Setting $$f'(t) = 0$$, we get $$\frac{1}{(1-t)^2} = \frac{4}{t^2}$$, so $$t^2 = 4(1-t)^2$$.

Taking the positive square root (since $$t > 0$$ and $$1 - t > 0$$), $$t = 2(1 - t)$$, which gives $$t = 2 - 2t$$, so $$3t = 2$$ and $$t = \frac{2}{3}$$.

Now $$f\left(\frac{2}{3}\right) = \frac{4}{2/3} + \frac{1}{1 - 2/3} = 6 + 3 = 9$$.

We verify this is a minimum by checking the second derivative or noting that $$f(t) \to \infty$$ as $$t \to 0^+$$ and $$t \to 1^-$$, so the critical point must be a minimum.

Hence, the answer is $$9$$.

We begin with the given trigonometric equation

$$ (k+1)\tan^2x-\sqrt{2}\,\lambda \tan x = (1-k), \qquad k\neq -1, \; \lambda\in\mathbb R. $$

Let us put $$t=\tan x.$$ Then the equation becomes a quadratic in $$t$$:

$$ (k+1)t^{2}-\sqrt2\,\lambda\,t-(1-k)=0. $$

The two real roots of this quadratic correspond to

$$ t_{1}= \tan\alpha,\qquad t_{2}= \tan\beta. $$

For a quadratic $$at^{2}+bt+c=0,$$ the standard Vieta relations state

$$\text{Sum of roots}= -\dfrac{b}{a},\qquad\text{Product of roots}= \dfrac{c}{a}.$$

Here we have $$a=k+1,\; b=-\sqrt2\,\lambda,\; c=-(1-k),$$ so

$$ S=t_{1}+t_{2}= -\frac{-\sqrt2\,\lambda}{\,k+1\,}= \frac{\sqrt2\,\lambda}{k+1}, $$

$$ P=t_{1}t_{2}= \frac{-(1-k)}{\,k+1\,}= \frac{k-1}{k+1}. $$

We are given the extra information

$$\tan^{2}(\alpha+\beta)=50.$$

Using the tangent-addition formula,

$$ \tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\,\tan\beta} =\frac{S}{1-P}. $$

Therefore

$$ \tan^{2}(\alpha+\beta)=\left(\frac{S}{1-P}\right)^{2}=50. $$

Substituting the expressions for $$S$$ and $$P$$ we get

$$ \left(\frac{\dfrac{\sqrt2\,\lambda}{k+1}}{\,1-\dfrac{k-1}{k+1}\,}\right)^{2}=50. $$

First calculate the denominator:

$$ 1-P = 1-\frac{k-1}{k+1} = \frac{k+1}{k+1}-\frac{k-1}{k+1} = \frac{(k+1)-(k-1)}{k+1} = \frac{2}{k+1}. $$

Hence

$$ \tan^{2}(\alpha+\beta)=\left(\frac{\sqrt2\,\lambda}{k+1}\,\frac{k+1}{2}\right)^{2} =\left(\frac{\sqrt2\,\lambda}{2}\right)^{2}=50. $$

Simplifying the square gives

$$ \left(\frac{\sqrt2\,\lambda}{2}\right)^{2} =\frac{2\lambda^{2}}{4} =\frac{\lambda^{2}}{2} =50. $$

Multiplying by 2,

$$ \lambda^{2}=100. $$

So

$$ \lambda=\pm 10. $$

The question asks for “a value of $$\lambda$$,” and among the options the only value that appears is $$10.$$ Therefore we select that.

Hence, the correct answer is Option B.

We have been given two infinite series

$$x \;=\; \sum_{n=0}^{\infty}(-1)^n \tan^{2n}\theta \qquad\text{and}\qquad y \;=\; \sum_{n=0}^{\infty}\cos^{2n}\theta,$$ with the restriction $$0<\theta<\dfrac{\pi}{4}.$$

Because $$0<\theta<\dfrac{\pi}{4}$$ we know $$\tan\theta<1$$ and $$\cos\theta<1,$$ so $$|\,-\tan^{2}\theta\,|<1$$ and $$|\cos^{2}\theta|<1.$$ Therefore both series are convergent geometric series, and we may apply the standard infinite-geometric-series formula.

Formula used. For an infinite geometric series with first term $$a$$ and common ratio $$r$$ satisfying $$|r|<1,$$ the sum is

$$S \;=\;\dfrac{a}{1-r}\,.$$

Finding $$x$$. For the series that defines $$x$$, the first term is obtained by putting $$n=0$$:

$$a_x \;=\;(-1)^0\tan^{0}\theta \;=\;1.$$

The common ratio is the factor that multiplies one term to obtain the next. Setting $$n\mapsto n+1$$ we see

$$r_x \;=\;-\,\tan^{2}\theta.$$

Applying the formula,

$$x \;=\;\dfrac{a_x}{1-r_x} \;=\;\dfrac{1}{1-(-\tan^{2}\theta)} \;=\;\dfrac{1}{1+\tan^{2}\theta}.$$

We now invoke the well-known trigonometric identity $$1+\tan^{2}\theta=\sec^{2}\theta.$$ Substituting,

$$x \;=\;\dfrac{1}{\sec^{2}\theta} \;=\;\cos^{2}\theta.$$

Finding $$y$$. For the series that defines $$y$$, the first term is obtained by putting $$n=0$$:

$$a_y \;=\;\cos^{0}\theta \;=\;1.$$

The common ratio is

$$r_y \;=\;\cos^{2}\theta.$$

Again using the formula,

$$y \;=\;\dfrac{a_y}{1-r_y} \;=\;\dfrac{1}{1-\cos^{2}\theta}.$$

From the Pythagorean identity $$\sin^{2}\theta +\cos^{2}\theta =1$$ we have $$1-\cos^{2}\theta=\sin^{2}\theta.$$ Substituting,

$$y \;=\;\dfrac{1}{\sin^{2}\theta} \;=\;\csc^{2}\theta.$$

Evaluating the required combinations. The option that interests us most is $$y(1-x).$$ Let us compute it step by step:

$$y(1-x) \;=\;\csc^{2}\theta\bigl(1-\cos^{2}\theta\bigr).$$

But we already know $$1-\cos^{2}\theta=\sin^{2}\theta.$$ Substituting,

$$y(1-x) \;=\;\csc^{2}\theta\,\sin^{2}\theta \;=\;\dfrac{1}{\sin^{2}\theta}\,\sin^{2}\theta \;=\;1.$$

Thus we have obtained the identity

$$y(1-x)=1.$$

No other option yields the value $$1$$, so the relationship demanded by the question corresponds precisely to option B.

Hence, the correct answer is Option B.

We have to evaluate

$$$\cos^{3}\!\left(\frac{\pi}{8}\right)\,\cos\!\left(\frac{3\pi}{8}\right)\;+\;\sin^{3}\!\left(\frac{\pi}{8}\right)\,\sin\!\left(\frac{3\pi}{8}\right).$$$

First, we recall the standard power-reduction formulas:

$$\cos^{3}\theta=\frac{1}{4}\bigl(3\cos\theta+\cos3\theta\bigr),$$

$$\sin^{3}\theta=\frac{1}{4}\bigl(3\sin\theta-\sin3\theta\bigr).$$

We apply these with $$\theta=\frac{\pi}{8}.$$ Substituting, we obtain

$$\cos^{3}\!\left(\frac{\pi}{8}\right)=\frac{1}{4}\Bigl(3\cos\!\left(\frac{\pi}{8}\right)+\cos\!\left(\frac{3\pi}{8}\right)\Bigr),$$

$$\sin^{3}\!\left(\frac{\pi}{8}\right)=\frac{1}{4}\Bigl(3\sin\!\left(\frac{\pi}{8}\right)-\sin\!\left(\frac{3\pi}{8}\right)\Bigr).$$

Now we multiply each of these by the corresponding cosine or sine factor present in the original expression:

$$$\cos^{3}\!\left(\frac{\pi}{8}\right)\cos\!\left(\frac{3\pi}{8}\right) =\frac{1}{4}\Bigl(3\cos\!\left(\frac{\pi}{8}\right)+\cos\!\left(\frac{3\pi}{8}\right)\Bigr) \cos\!\left(\frac{3\pi}{8}\right),$$$

$$$\sin^{3}\!\left(\frac{\pi}{8}\right)\sin\!\left(\frac{3\pi}{8}\right) =\frac{1}{4}\Bigl(3\sin\!\left(\frac{\pi}{8}\right)-\sin\!\left(\frac{3\pi}{8}\right)\Bigr) \sin\!\left(\frac{3\pi}{8}\right).$$$

Adding the two products gives

$$$\begin{aligned} E &= \frac{1}{4}\Bigl(3\cos\!\left(\frac{\pi}{8}\right)+\cos\!\left(\frac{3\pi}{8}\right)\Bigr) \cos\!\left(\frac{3\pi}{8}\right)\\[4pt] &\quad+\frac{1}{4}\Bigl(3\sin\!\left(\frac{\pi}{8}\right)-\sin\!\left(\frac{3\pi}{8}\right)\Bigr) \sin\!\left(\frac{3\pi}{8}\right)\\[6pt] &=\frac{1}{4}\Bigl[ 3\cos\!\left(\frac{\pi}{8}\right)\cos\!\left(\frac{3\pi}{8}\right) +\cos^{2}\!\left(\frac{3\pi}{8}\right) +3\sin\!\left(\frac{\pi}{8}\right)\sin\!\left(\frac{3\pi}{8}\right) -\sin^{2}\!\left(\frac{3\pi}{8}\right)\Bigr]. \end{aligned}$$$

We notice in the bracket two distinct parts. For the terms with the coefficient 3 we use the angle-difference identity

$$\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos(\alpha-\beta).$$

Taking $$\alpha=\frac{3\pi}{8},\;\beta=\frac{\pi}{8},$$ we get

$$$\cos\!\left(\frac{3\pi}{8}\right)\cos\!\left(\frac{\pi}{8}\right) +\sin\!\left(\frac{3\pi}{8}\right)\sin\!\left(\frac{\pi}{8}\right) =\cos\!\left(\frac{3\pi}{8}-\frac{\pi}{8}\right) =\cos\!\left(\frac{2\pi}{8}\right) =\cos\!\left(\frac{\pi}{4}\right).$$$

So the portion with the coefficient 3 simplifies to

$$3\cos\!\left(\frac{\pi}{4}\right).$$

Next, the remaining pair $$\cos^{2}\!\left(\frac{3\pi}{8}\right)-\sin^{2}\!\left(\frac{3\pi}{8}\right)$$ can be recognized through the double-angle identity

$$\cos^{2}\theta-\sin^{2}\theta=\cos2\theta.$$

Putting $$\theta=\frac{3\pi}{8},$$ we have

$$$\cos^{2}\!\left(\frac{3\pi}{8}\right)-\sin^{2}\!\left(\frac{3\pi}{8}\right) =\cos\!\left(2\cdot\frac{3\pi}{8}\right) =\cos\!\left(\frac{6\pi}{8}\right) =\cos\!\left(\frac{3\pi}{4}\right).$$$

Collecting these results, the bracketed expression becomes

$$3\cos\!\left(\frac{\pi}{4}\right)+\cos\!\left(\frac{3\pi}{4}\right).$$

Therefore,

$$E=\frac{1}{4}\Bigl[3\cos\!\left(\frac{\pi}{4}\right)+\cos\!\left(\frac{3\pi}{4}\right)\Bigr].$$

We now substitute the numeric values:

$$\cos\!\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt2},\qquad \cos\!\left(\frac{3\pi}{4}\right)=-\frac{1}{\sqrt2}.$$

Thus,

$$$E=\frac{1}{4}\Bigl[3\!\left(\frac{1}{\sqrt2}\right)+\!\left(-\frac{1}{\sqrt2}\right)\Bigr] =\frac{1}{4}\!\left(\frac{3-1}{\sqrt2}\right) =\frac{1}{4}\!\left(\frac{2}{\sqrt2}\right) =\frac{2}{4\sqrt2} =\frac{1}{2\sqrt2}.$$$

Hence, the correct answer is Option B.

We have two expressions, namely $$L = \sin^{2}\!\left(\dfrac{\pi}{16}\right) - \sin^{2}\!\left(\dfrac{\pi}{8}\right)$$ and $$M = \cos^{2}\!\left(\dfrac{\pi}{16}\right) - \sin^{2}\!\left(\dfrac{\pi}{8}\right).$$ Our task is to simplify each of them and then match the results with the options given.

First we handle $$L$$. We recall the double-angle form for squares of sine: the identity $$\sin^{2}\theta = \dfrac{1 - \cos 2\theta}{2}.$$ Applying this identity to both terms we get

$$\sin^{2}\!\left(\dfrac{\pi}{16}\right) = \dfrac{1 - \cos\!\left(\dfrac{\pi}{8}\right)}{2},$$

$$\sin^{2}\!\left(\dfrac{\pi}{8}\right) = \dfrac{1 - \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

Substituting these forms into $$L$$ we obtain

$$L = \dfrac{1 - \cos\!\left(\dfrac{\pi}{8}\right)}{2} \;-\; \dfrac{1 - \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

The two fractions have the same denominator, so we can combine the numerators directly:

$$L = \dfrac{1 - \cos\!\left(\dfrac{\pi}{8}\right) - 1 + \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

Inside the numerator the two “1” terms cancel, giving

$$L = \dfrac{-\cos\!\left(\dfrac{\pi}{8}\right) + \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

We know that $$\cos\!\left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}.$$ Substituting this numerical value we arrive at

$$L = \dfrac{-\cos\!\left(\dfrac{\pi}{8}\right) + \dfrac{1}{\sqrt{2}}}{2} = -\dfrac{1}{2}\cos\!\left(\dfrac{\pi}{8}\right) + \dfrac{1}{2\sqrt{2}}.$$

So we have simplified $$L$$ to the form $$L = -\dfrac{1}{2}\cos\!\left(\dfrac{\pi}{8}\right) + \dfrac{1}{2\sqrt{2}}.$$ We keep this result in mind when we compare with the listed choices.

Now we simplify $$M$$. For this we use the companion identity for cosine squares, namely $$\cos^{2}\theta = \dfrac{1 + \cos 2\theta}{2}.$$ Applying it to the first term of $$M$$ yields

$$\cos^{2}\!\left(\dfrac{\pi}{16}\right) = \dfrac{1 + \cos\!\left(\dfrac{\pi}{8}\right)}{2}.$$

The second term $$\sin^{2}\!\left(\dfrac{\pi}{8}\right)$$ we have already expressed above, viz.

$$\sin^{2}\!\left(\dfrac{\pi}{8}\right) = \dfrac{1 - \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

Substituting both of these into the definition of $$M$$ we get

$$M = \dfrac{1 + \cos\!\left(\dfrac{\pi}{8}\right)}{2} \;-\; \dfrac{1 - \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

Combining the numerators over the common denominator gives

$$M = \dfrac{1 + \cos\!\left(\dfrac{\pi}{8}\right) - 1 + \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

The two “1” terms again cancel, leaving

$$M = \dfrac{\cos\!\left(\dfrac{\pi}{8}\right) + \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

Substituting $$\cos\!\left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$$ we obtain

$$M = \dfrac{\cos\!\left(\dfrac{\pi}{8}\right) + \dfrac{1}{\sqrt{2}}}{2} = \dfrac{1}{2}\cos\!\left(\dfrac{\pi}{8}\right) + \dfrac{1}{2\sqrt{2}}.$$

We now compare the simplified forms with the four options furnished in the question.

The expression derived for $$L$$ is $$-\dfrac{1}{2}\cos\!\left(\dfrac{\pi}{8}\right) + \dfrac{1}{2\sqrt{2}},$$ which does not match either Option A or Option B. On the other hand, the expression derived for $$M$$ is $$\dfrac{1}{2\sqrt{2}} + \dfrac{1}{2}\cos\!\left(\dfrac{\pi}{8}\right),$$ and this coincides exactly with Option D.

Hence, the correct answer is Option D.

We begin with the given equation

$$\cos^4\theta+\sin^4\theta+\lambda=0.$$

For real values of $$\theta$$ to exist, the trigonometric part $$\cos^4\theta+\sin^4\theta$$ must be equal to $$-\lambda$$ for some $$\theta$$. So we first find the range of the expression

$$f(\theta)=\cos^4\theta+\sin^4\theta.$$

Write $$\cos^4\theta+\sin^4\theta$$ in terms of $$\cos^2\theta$$ and $$\sin^2\theta$$. Let

$$a=\cos^2\theta,\qquad b=\sin^2\theta.$$

We know the Pythagorean identity $$a+b=1$$ and, of course, $$0\le a\le1,\;0\le b\le1.$$ Using the algebraic identity $$a^2+b^2=(a+b)^2-2ab,$$ we have

$$\cos^4\theta+\sin^4\theta =a^2+b^2 =(a+b)^2-2ab =1^2-2ab =1-2\cos^2\theta\sin^2\theta.$$

Next, recall the double-angle formula $$\sin2\theta=2\sin\theta\cos\theta,$$ so that

$$\sin^22\theta=4\sin^2\theta\cos^2\theta \;\Longrightarrow\; \cos^2\theta\sin^2\theta=\dfrac{\sin^22\theta}{4}.$$

Substituting this into the previous expression gives

$$f(\theta)=1-2\left(\dfrac{\sin^22\theta}{4}\right) =1-\dfrac{1}{2}\sin^22\theta.$$

The term $$\sin^22\theta$$ always lies in the interval $$[0,1]$$ because the square of a sine function never exceeds 1 and is never negative. Hence

$$-\dfrac12\le -\dfrac{1}{2}\sin^22\theta\le0,$$ and adding 1 to every part we obtain the range of $$f(\theta):$$

$$\boxed{\dfrac12\le f(\theta)\le1}.$$

This means the expression $$\cos^4\theta+\sin^4\theta$$ can take any value from $$\dfrac12$$ up to $$1$$, inclusive.

Now the original equation can be written as

$$f(\theta)+\lambda=0 \;\Longrightarrow\; f(\theta)=-\lambda.$$

Thus, for some real $$\theta,$$ the number $$-\lambda$$ must lie in the interval $$\left[\dfrac12,1\right]$$ determined above. Multiplying that interval by $$-1$$ reverses the inequality signs, giving the permissible interval for $$\lambda:$$

$$-\left[\,\dfrac12,1\,\right]=\left[-1,-\dfrac12\right].$$

We include the endpoints because both $$\sin^22\theta=0$$ (giving $$f(\theta)=1$$) and $$\sin^22\theta=1$$ (giving $$f(\theta)=\dfrac12$$) are achievable for suitable values of $$\theta$$ (for example, $$\theta=0$$ and $$\theta=\dfrac{\pi}{4}$$ respectively).

Therefore

$$\boxed{\lambda\in\left[-1,-\dfrac12\right]}.$$

Among the given choices, this corresponds exactly to Option B.

Hence, the correct answer is Option B.

Let the still surface of the lake be a horizontal plane. Choose a point $$A$$ on this surface such that the given point $$P$$ lies vertically above $$A$$. We are told that $$P$$ is 200 m above the lake, so $$PA = 200\text{ m}$$.

Let the cloud be $$C$$ and its image in the lake be $$C'$$. Because an image in a perfectly still lake is formed by reflection in the water surface, $$C$$ and $$C'$$ are on the same vertical line through $$A$$, with $$A$$ exactly midway between them. If we denote the height of the cloud above the lake by $$h$$ metres, then

$$AC = h,\qquad AC' = h,\qquad \text{and}\qquad C'$$ is $$h$$ metres below the water surface.

Put the horizontal distance of both $$C$$ and $$C'$$ from the foot $$A$$ as $$x$$ metres. Thus the coordinates (taking the lake surface as the $$x$$-axis and up as positive $$y$$) can be visualised as

$$A(0,0),\; P(0,200),\; C(x,h),\; C'(x,-h).$$

The angle of elevation of $$C$$ from $$P$$ is 30°. By the definition of tangent in a right triangle,

$$\tan 30^{\circ} \;=\; \frac{\text{opposite side}}{\text{adjacent side}} \;=\; \frac{(h-200)}{x}.$$

Since $$\tan 30^{\circ} = \frac{1}{\sqrt{3}}$$, we get

$$\frac{1}{\sqrt{3}} = \frac{h-200}{x} \;\;\Longrightarrow\;\; x = \sqrt{3}\,(h-200). \quad (1)$$

Next, the angle of depression of the image $$C'$$ from $$P$$ is 60°. The angle of depression equals the angle the line of sight makes below the horizontal. Using the same tangent definition, we have

$$\tan 60^{\circ} = \frac{\text{vertical drop from }P\text{ to }C'}{\text{horizontal distance }x} = \frac{200+h}{x}.$$

Because $$\tan 60^{\circ} = \sqrt{3}$$, this yields

$$\sqrt{3} = \frac{200+h}{x} \;\;\Longrightarrow\;\; x = \frac{200+h}{\sqrt{3}}. \quad (2)$$

Both equations (1) and (2) represent the same value of $$x$$, so we equate them:

$$\sqrt{3}(h-200) = \frac{200+h}{\sqrt{3}}.$$

Multiplying both sides by $$\sqrt{3}$$ to clear the denominator,

$$3(h-200) = 200 + h.$$

Expanding and rearranging step by step,

$$3h - 600 = 200 + h$$
$$3h - h = 200 + 600$$
$$2h = 800$$
$$h = 400.$$ So the cloud is $$400\text{ m}$$ above the lake.

The vertical separation between $$P$$ and $$C$$ is therefore

$$\text{vertical difference} = h - 200 = 400 - 200 = 200\text{ m}.$$

Substituting $$h = 400$$ into equation (1) (or equation (2)) to find $$x$$,

$$x = \sqrt{3}(h-200) = \sqrt{3}\times 200 = 200\sqrt{3}\text{ m}.$$

Now we know both the horizontal and vertical components of the line segment $$PC$$. Using the Pythagoras theorem, which states $$\text{hypotenuse}^2 = \text{base}^2 + \text{perpendicular}^2,$$ we have

$$PC = \sqrt{(200)^2 + (200\sqrt{3})^2} = \sqrt{40000 + 120000} = \sqrt{160000} = 400\text{ m}.$$

Hence, the correct answer is Option C.

Let us denote the foot of the mountain by point $$B$$, the first observation point on the ground by $$A$$ and the summit of the mountain by $$S$$. We are asked to find the height $$BS$$ (in km) of the mountain.

We first translate the given angles into the language of right-angle trigonometry. From point $$A$$ the angle of elevation of the summit is $$45^\circ$$. Using the definition of the tangent function,

$$\tan 45^\circ \;=\;\frac{\text{opposite side}}{\text{adjacent side}} \;=\;\frac{BS}{AB}.$$

Because $$\tan 45^\circ = 1$$, we obtain the simple relation

$$BS = AB.$$

We now move from $$A$$ a distance of exactly $$1\text{ km}$$ towards the summit along a path that is inclined at $$30^\circ$$ to the horizontal. Let the new point reached on this path be $$C$$. Since the traveller is going “towards the summit”, the path $$AC$$ lies in the plane containing $$AB$$ and $$BS$$. From elementary resolution of a line segment inclined at $$30^\circ$$ we get:

vertical rise from $$A$$ to $$C$$: $$AC \sin 30^\circ = 1 \times \frac12 = 0.5\text{ km},$$

horizontal advance (towards the foot $$B$$): $$AC \cos 30^\circ = 1 \times \frac{\sqrt3}{2} = \frac{\sqrt3}{2}\text{ km}.$$

Hence the height of point $$C$$ above the ground is

$$\text{height of }C = 0.5\text{ km},$$

and the remaining vertical distance from $$C$$ to the summit is

$$SC = BS - 0.5\text{ km}.$$

Meanwhile the horizontal distance from point $$C$$ to the foot $$B$$ has decreased by $$\dfrac{\sqrt3}{2}\text{ km}$$, so it is now

$$BC = AB - \frac{\sqrt3}{2}\text{ km}.$$

But we already know $$AB = BS$$, so

$$BC = BS - \frac{\sqrt3}{2}.$$

From point $$C$$ the angle of elevation of the summit is stated to be $$60^\circ$$. Applying the tangent definition once more, this time in right-angled triangle $$B C S$$, we have

$$\tan 60^\circ = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{SC}{BC} = \frac{BS - 0.5}{\,BS - \dfrac{\sqrt3}{2}\,}.$$

Because $$\tan 60^\circ = \sqrt3$$, we write

$$\sqrt3 = \frac{BS - 0.5}{BS - \dfrac{\sqrt3}{2}}.$$

To remove the denominator, we cross-multiply:

$$\sqrt3\Bigl(BS - \frac{\sqrt3}{2}\Bigr) = BS - 0.5.$$

Expanding the left-hand side gives

$$\sqrt3\,BS - \sqrt3\!\left(\frac{\sqrt3}{2}\right) = BS - 0.5,$$

and since $$\sqrt3 \times \sqrt3 = 3$$, this becomes

$$\sqrt3\,BS - \frac{3}{2} = BS - 0.5.$$

Now we gather the $$BS$$ terms on one side and the constants on the other side:

$$\sqrt3\,BS - BS = \frac{3}{2} - 0.5.$$

The right-hand constant simplifies because $$0.5 = \dfrac{1}{2}$$, so

$$\sqrt3\,BS - BS = \frac{3}{2} - \frac{1}{2} = 1.$$

Factoring $$BS$$ out of the left-hand side, we obtain

$$(\sqrt3 - 1)\,BS = 1.$$

Finally we solve for $$BS$$ (the required height):

$$BS = \frac{1}{\sqrt3 - 1}.$$

This expression exactly matches Option C given in the problem statement. Hence, the correct answer is Option C.

We want the least possible value of the function

$$f(x)=2^{\sin x}+2^{\cos x},\qquad x\in\mathbb R.$$

Because the expression is periodic with period $$2\pi,$$ it is enough to study one full period, for example $$x\in[0,2\pi).$$ To locate extrema we differentiate. The basic differentiation formula we need is

$$\dfrac{d}{dx}\bigl(2^{u(x)}\bigr)=2^{u(x)}\ln 2\;u'(x).$$

Applying it first to $$2^{\sin x}$$ and then to $$2^{\cos x}$$ gives

$$\frac{d}{dx}\bigl(2^{\sin x}\bigr)=2^{\sin x}\ln 2\;\cos x,$$

$$\frac{d}{dx}\bigl(2^{\cos x}\bigr)=2^{\cos x}\ln 2\;(-\sin x).$$

So the derivative of the whole function is

$$f'(x)=2^{\sin x}\ln 2\;\cos x-2^{\cos x}\ln 2\;\sin x.$$

Factorising the common factor $$\ln 2$$ (which is always positive) we have

$$f'(x)=\ln 2\Bigl(\cos x\;2^{\sin x}-\sin x\;2^{\cos x}\Bigr).$$

The critical points occur when the bracket equals zero, i.e.

$$\cos x\;2^{\sin x}-\sin x\;2^{\cos x}=0.$$

Re-arranging gives the necessary condition

$$\cos x\;2^{\sin x}=\sin x\;2^{\cos x}.$$

Assuming $$\sin x\neq 0$$ and $$\cos x\neq 0$$ (we shall look at those special cases later) we divide by the positive quantities and obtain

$$\frac{\cos x}{\sin x}=2^{\cos x-\sin x}.$$

The left side is $$\cot x,$$ the right side is $$2^{\cos x-\sin x}.$$ This transcendental equation is satisfied when the two exponents of 2 are equal, that is when

$$\sin x=\cos x.$$

Using the identity $$\sin x=\cos x\; \Longrightarrow\; x=\frac{\pi}{4}+k\pi,\;k\in\mathbb Z,$$ we get two types of points in one period:

1. $$x=\frac{\pi}{4}$$ (both sine and cosine positive),
2. $$x=\frac{5\pi}{4}$$ (both sine and cosine negative).

Next we check the endpoints where either sine or cosine is zero, because there the derivative formula involved division by those terms:

a) $$x=0:\;\sin x=0,\;\cos x=1\;\Longrightarrow\;f(0)=2^{0}+2^{1}=1+2=3.$$

b) $$x=\frac{\pi}{2}:\;\sin x=1,\;\cos x=0\;\Longrightarrow\;f\!\left(\frac{\pi}{2}\right)=2^{1}+2^{0}=2+1=3.$$

c) $$x=\pi:\;\sin x=0,\;\cos x=-1\;\Longrightarrow\;f(\pi)=2^{0}+2^{-1}=1+\frac12=\frac32=1.5.$$

d) $$x=\frac{3\pi}{2}:\;\sin x=-1,\;\cos x=0\;\Longrightarrow\;f\!\left(\frac{3\pi}{2}\right)=2^{-1}+2^{0}=\frac12+1=1.5.$$

Now we evaluate the function at the two critical points where $$\sin x=\cos x.$$

For $$x=\frac{\pi}{4}$$ we have $$\sin x=\cos x=\frac1{\sqrt2},$$ so

$$f\!\left(\frac{\pi}{4}\right)=2^{\frac1{\sqrt2}}+2^{\frac1{\sqrt2}}=2\cdot2^{\frac1{\sqrt2}}=2^{1+\frac1{\sqrt2}}\approx3.266.$$

For $$x=\frac{5\pi}{4}$$ we have $$\sin x=\cos x=-\frac1{\sqrt2},$$ so

$$f\!\left(\frac{5\pi}{4}\right)=2^{-\frac1{\sqrt2}}+2^{-\frac1{\sqrt2}}=2\cdot2^{-\frac1{\sqrt2}}=2^{1-\frac1{\sqrt2}}\approx1.225.$$

Comparing all the values we have collected

$$\begin{aligned} f\!\left(\frac{5\pi}{4}\right)&=2^{1-\frac1{\sqrt2}}\approx1.225,\\[4pt] f(\pi)=f\!\left(\frac{3\pi}{2}\right)&=1.5,\\[4pt] f(0)=f\!\left(\frac{\pi}{2}\right)&=3,\\[4pt] f\!\left(\frac{\pi}{4}\right)&\approx3.266, \end{aligned}$$

and note that all other points in the interval $$[0,2\pi)$$ lie between these tested points. Hence the smallest of all these numbers, and therefore the global minimum of $$f(x),$$ is

$$f_{\min}=2^{1-\frac1{\sqrt2}}.$$

Thus the minimum value of $$2^{\sin x}+2^{\cos x}$$ is $$2^{1-\frac1{\sqrt2}}.$$

Hence, the correct answer is Option D.

We have to solve the trigonometric equation $$2\cot^2\theta-\dfrac{5}{\sin\theta}+4=0$$ for $$\theta\in(0,2\pi)\setminus\{\pi\}$$, obtain the smallest root $$\theta_1$$ and the largest root $$\theta_2$$ in that interval, and then evaluate the definite integral $$\displaystyle\int_{\theta_1}^{\theta_2}\cos^2 3\theta\,d\theta$$.

First, recall that $$\cot\theta=\dfrac{\cos\theta}{\sin\theta}$$, so $$\cot^2\theta=\dfrac{\cos^2\theta}{\sin^2\theta}$$. Substituting this in the given equation we get

$$2\left(\dfrac{\cos^2\theta}{\sin^2\theta}\right)-\dfrac{5}{\sin\theta}+4=0.$$

To clear denominators we multiply every term by $$\sin^2\theta$$:

$$2\cos^2\theta-5\sin\theta+4\sin^2\theta=0.$$

Now we express $$\cos^2\theta$$ in terms of $$\sin\theta$$ by using the Pythagorean identity $$\cos^2\theta=1-\sin^2\theta$$. Let $$x=\sin\theta$$ for simplicity. Substituting $$\cos^2\theta=1-x^2$$ gives

$$2(1-x^2)-5x+4x^2=0.$$

Expanding and collecting like terms:

$$2-2x^2-5x+4x^2=0,$$

$$2+2x^2-5x=0.$$

Re-ordering, we get the quadratic equation

$$2x^2-5x+2=0.$$

To find its roots we use the quadratic formula. For $$ax^2+bx+c=0$$, the roots are $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$. Here $$a=2,\;b=-5,\;c=2$$, so

$$x=\dfrac{5\pm\sqrt{(-5)^2-4\cdot2\cdot2}}{2\cdot2} =\dfrac{5\pm\sqrt{25-16}}{4} =\dfrac{5\pm\sqrt{9}}{4} =\dfrac{5\pm3}{4}.$$

This gives two numerical roots:

$$x_1=\dfrac{5+3}{4}=2,\qquad x_2=\dfrac{5-3}{4}=\dfrac12.$$

Since $$x=\sin\theta$$ must satisfy $$-1\le x\le1$$, the value $$x=2$$ is impossible. Thus the only admissible solution is

$$\sin\theta=\dfrac12.$$

Within the interval $$(0,2\pi)$$ (excluding $$\pi$$) the equation $$\sin\theta=\dfrac12$$ is satisfied at

$$\theta=\dfrac\pi6\quad\text{and}\quad\theta=\dfrac{5\pi}6.$$

There are no other values in $$\bigl(\pi,2\pi\bigr)$$ with $$\sin\theta=\dfrac12$$ because $$\sin\theta$$ becomes negative beyond $$\pi$$. Therefore,

$$\theta_1=\dfrac{\pi}{6},\qquad \theta_2=\dfrac{5\pi}{6}.$$

Now we have to evaluate the definite integral

$$\int_{\theta_1}^{\theta_2}\cos^2 3\theta\,d\theta =\int_{\pi/6}^{5\pi/6}\cos^2 3\theta\,d\theta.$$

We use the power-reduction identity for the square of cosine:

$$\cos^2\alpha=\dfrac{1+\cos2\alpha}{2}.$$

With $$\alpha=3\theta$$, this becomes

$$\cos^2 3\theta=\dfrac{1+\cos6\theta}{2}.$$

Substituting into the integral gives

$$\int_{\pi/6}^{5\pi/6}\cos^2 3\theta\,d\theta =\int_{\pi/6}^{5\pi/6}\dfrac{1+\cos6\theta}{2}\,d\theta.$$

We can split and integrate term by term:

$$\int_{\pi/6}^{5\pi/6}\dfrac{1}{2}\,d\theta+\int_{\pi/6}^{5\pi/6}\dfrac{\cos6\theta}{2}\,d\theta =\dfrac12\int_{\pi/6}^{5\pi/6}d\theta+\dfrac12\int_{\pi/6}^{5\pi/6}\cos6\theta\,d\theta.$$

The first integral is straightforward:

$$\dfrac12\int_{\pi/6}^{5\pi/6}d\theta=\dfrac12\bigl[\theta\bigr]_{\pi/6}^{5\pi/6} =\dfrac12\left(\dfrac{5\pi}{6}-\dfrac{\pi}{6}\right) =\dfrac12\cdot\dfrac{4\pi}{6} =\dfrac{4\pi}{12} =\dfrac{\pi}{3}.$$

For the second integral we use the standard antiderivative $$\int\cos k\theta\,d\theta=\dfrac{\sin k\theta}{k}.$$ Here $$k=6,$$ so

$$\int\cos6\theta\,d\theta=\dfrac{\sin6\theta}{6}.$$

Therefore,

$$\dfrac12\int_{\pi/6}^{5\pi/6}\cos6\theta\,d\theta =\dfrac12\left[\dfrac{\sin6\theta}{6}\right]_{\pi/6}^{5\pi/6} =\dfrac{1}{12}\bigl[\sin6\theta\bigr]_{\pi/6}^{5\pi/6}.$$

Now we evaluate $$\sin6\theta$$ at the limits:

At $$\theta=\dfrac{5\pi}{6}:$$ $$6\theta=5\pi,$$ so $$\sin5\pi=0.$$

At $$\theta=\dfrac{\pi}{6}:$$ $$6\theta=\pi,$$ so $$\sin\pi=0.$$

Hence

$$\dfrac{1}{12}\bigl[\sin6\theta\bigr]_{\pi/6}^{5\pi/6} =\dfrac{1}{12}(0-0)=0.$$

Adding the two parts, the value of the integral is

$$\dfrac{\pi}{3}+0=\dfrac{\pi}{3}.$$

Hence, the correct answer is Option A.

Let the foot of the hill be the point $$B$$ and let the top of the hill be the point $$T$$. We take the horizontal plane through $$B$$ as our reference ground level.

We first stand at a point $$P$$ on this horizontal plane such that the angle of elevation of $$T$$ from $$P$$ is $$45^\circ$$. If the horizontal distance $$BP$$ is denoted by $$x$$ metres and the vertical height of the hill $$BT$$ is denoted by $$h$$ metres, then by the definition of tangent we have

$$\tan 45^\circ=\frac{\text{opposite}}{\text{adjacent}}=\frac{h}{x}.$$

Since $$\tan 45^\circ=1$$, this at once gives

$$h=x.$$

Now we walk from $$P$$ directly towards the top, along a path that is inclined at $$30^\circ$$ to the horizontal, and we cover a distance of $$80$$ metres along this path. Let the new point of observation be $$Q$$. Because the path makes an angle of $$30^\circ$$ with the horizontal, we can resolve the $$80$$ m into horizontal and vertical components using the standard relations

$$\text{horizontal component}=80\cos30^\circ,\qquad\text{vertical component}=80\sin30^\circ.$$

Recalling that $$\cos30^\circ=\dfrac{\sqrt3}{2}$$ and $$\sin30^\circ=\dfrac12$$, we obtain

$$\text{horizontal component}=80\left(\frac{\sqrt3}{2}\right)=40\sqrt3,$$ $$\text{vertical component}=80\left(\frac12\right)=40.$$

Thus, compared with point $$P$$, the point $$Q$$ is $$40\sqrt3$$ m closer to the foot $$B$$ horizontally and $$40$$ m higher vertically. Consequently the horizontal distance from $$Q$$ to $$B$$ equals

$$x-40\sqrt3,$$

and the height of $$Q$$ above the ground plane equals

$$40\text{ m}.$$

From $$Q$$ the angle of elevation of the top $$T$$ is now $$75^\circ$$. The vertical separation between $$Q$$ and $$T$$ is

$$h-40,$$

while the horizontal separation is

$$x-40\sqrt3.$$

Applying the definition of tangent again, we write

$$\tan75^\circ=\frac{h-40}{\,x-40\sqrt3\,}.$$

But we already have $$h=x$$ from the first observation, so we substitute $$x=h$$ to get

$$\tan75^\circ=\frac{h-40}{\,h-40\sqrt3\,}.$$

To solve for $$h$$ we first recall the exact value of $$\tan75^\circ$$. Using the angle-addition formula

$$\tan(45^\circ+30^\circ)=\frac{\tan45^\circ+\tan30^\circ}{1-\tan45^\circ\tan30^\circ},$$

and the facts $$\tan45^\circ=1$$ and $$\tan30^\circ=\dfrac1{\sqrt3},$$ we find

$$\tan75^\circ=\frac{1+\dfrac1{\sqrt3}}{1-1\cdot\dfrac1{\sqrt3}} =\frac{\sqrt3+1}{\sqrt3-1}.$$

Introduce the shorthand $$t=\tan75^\circ=\dfrac{\sqrt3+1}{\sqrt3-1}.$$ Our equation becomes

$$\frac{h-40}{h-40\sqrt3}=t.$$

Cross-multiplying,

$$h-40=t(h-40\sqrt3).$$

Expand the right side:

$$h-40=th-40t\sqrt3.$$

Rearrange to collect the $$h$$ terms on the left and constants on the right:

$$h-th=40-40t\sqrt3.$$

Factor out $$h$$ on the left:

$$h(1-t)=40(1-t\sqrt3).$$

Finally, divide by the coefficient $$1-t$$ (noting that $$t\neq1$$) to obtain $$h$$ explicitly:

$$h=40\frac{1-t\sqrt3}{1-t}.$$

We now substitute the exact value of $$t$$. First compute $$1-t$$ and $$1-t\sqrt3$$ separately:

$$1-t=1-\frac{\sqrt3+1}{\sqrt3-1} =\frac{\sqrt3-1-(\sqrt3+1)}{\sqrt3-1} =\frac{-2}{\sqrt3-1},$$

$$1-t\sqrt3=1-\sqrt3\left(\frac{\sqrt3+1}{\sqrt3-1}\right) =\frac{\sqrt3-1-(3+\sqrt3)}{\sqrt3-1} =\frac{-4}{\sqrt3-1}.$$

Taking the ratio,

$$\frac{1-t\sqrt3}{1-t} =\frac{\dfrac{-4}{\sqrt3-1}}{\dfrac{-2}{\sqrt3-1}} =\frac{-4}{-2}=2.$$

Therefore

$$h=40\times2=80.$$

Thus the height of the hill is $$80\text{ metres}$$.

So, the answer is $$80$$.

We start with the given transcendental equation

$$\log_{\frac12}\!\bigl|\sin x\bigr| \;=\; 2 \;-\; \log_{\frac12}\!\bigl|\cos x\bigr|$$

in the interval $$[0,\,2\pi]$$. Here the base of every logarithm is $$\dfrac12,$$ which lies between 0 and 1, so all logarithms are well-defined only when their arguments are positive; that already guarantees $$\sin x\neq0$$ and $$\cos x\neq0$$ for each admissible solution.

First we transpose the term $$-\log_{\frac12}\!\bigl|\cos x\bigr|$$ to the left side:

$$\log_{\frac12}\!\bigl|\sin x\bigr| \;+\; \log_{\frac12}\!\bigl|\cos x\bigr| \;=\; 2.$$

Now we invoke the well-known product rule for logarithms, stated as

$$\log_b m + \log_b n \;=\; \log_b(mn)\quad\text{for the same base }b.$$

Applying this rule with $$b=\dfrac12,\;m=|\sin x|,\;n=|\cos x|,$$ we obtain

$$\log_{\frac12}\!\bigl(|\sin x|\,|\cos x|\bigr) \;=\; 2.$$

Next we convert this logarithmic statement to its equivalent exponential form. For any base $$b\;(0<b\neq1)$$ we have the definition

$$\log_b A = C \;\Longleftrightarrow\; A = b^{\,C}.$$

Hence

$$|\sin x|\,|\cos x| \;=\; \Bigl(\tfrac12\Bigr)^{2} \;=\; \tfrac14.$$

Because the left-hand side contains the product of absolute values, we rewrite it through the double‐angle identity. We recall the formula

$$\sin 2x \;=\; 2\sin x\cos x,$$

and from it we find

$$|\sin x|\,|\cos x| \;=\; \tfrac12\,|2\sin x\cos x| \;=\; \tfrac12\,|\sin 2x|.$$

Substituting this expression into the equation $$|\sin x|\,|\cos x|=\tfrac14$$ gives

$$\tfrac12\,|\sin 2x| \;=\; \tfrac14,$$

and multiplying both sides by 2 yields the simpler condition

$$|\sin 2x| \;=\; \tfrac12.$$

Thus our task reduces to solving

$$\bigl|\sin 2x\bigr| = \frac12$$

for $$x$$ in the original domain $$[0,\,2\pi].$$ Set $$\theta = 2x;$$ then $$\theta$$ runs through the interval $$[0,\,4\pi]$$ while $$x$$ runs through $$[0,\,2\pi].$$ Therefore we must find all $$\theta\in[0,\,4\pi]$$ such that

$$|\sin\theta| = \frac12.$$

We know that $$\sin\theta = \pm\dfrac12$$ at the angles

$$\theta = \frac{\pi}{6},\;\frac{5\pi}{6},\;\frac{7\pi}{6},\;\frac{11\pi}{6}$$

within the first full cycle $$[0,\,2\pi].$$ Since the sine function has period $$2\pi,$$ the same four angles recur once more when we add $$2\pi,$$ giving in the second cycle

$$\theta = \frac{\pi}{6}+2\pi,\;\frac{5\pi}{6}+2\pi,\;\frac{7\pi}{6}+2\pi,\;\frac{11\pi}{6}+2\pi.$$

Altogether, in the whole interval $$[0,\,4\pi]$$ we therefore obtain eight distinct solutions for $$\theta.$$ Explicitly, they are

$$\theta = \frac{\pi}{6},\;\frac{5\pi}{6},\;\frac{7\pi}{6},\;\frac{11\pi}{6},\; \frac{13\pi}{6},\;\frac{17\pi}{6},\;\frac{19\pi}{6},\;\frac{23\pi}{6}.$$

Since $$\theta = 2x,$$ each such $$\theta$$ corresponds to a unique $$x=\dfrac{\theta}{2}$$ lying in $$[0,\,2\pi].$$ None of these $$x$$-values makes $$\sin x$$ or $$\cos x$$ equal to zero, so all are admissible for the original logarithmic equation. Consequently there are exactly eight distinct $$x$$ satisfying the given equation in the stated interval.

So, the answer is $$8$$.

We have been given two separate relations and both angles lie in the first quadrant, that is $$0<\alpha<\dfrac{\pi}{2}$$ and $$0<\beta<\dfrac{\pi}{2}$$, so all trigonometric ratios of these angles are positive.

First of all, consider the equation

$$\dfrac{\sqrt{2}\,\sin\alpha}{\sqrt{1+\cos 2\alpha}}=\dfrac{1}{7}.$$

We recall the double-angle identity

$$\cos 2\alpha = 2\cos^2\alpha - 1,$$

which can be rearranged to give

$$1 + \cos 2\alpha = 2\cos^2\alpha.$$

Substituting this into the denominator we obtain

$$\sqrt{1+\cos 2\alpha} = \sqrt{2\cos^2\alpha} = \sqrt{2}\,\cos\alpha,$$

because $$\cos\alpha>0$$ in the given interval and therefore the absolute value is not required.

Now we rewrite the left‐hand side of the original equation:

$$\dfrac{\sqrt{2}\,\sin\alpha}{\sqrt{2}\,\cos\alpha} = \dfrac{\sin\alpha}{\cos\alpha} = \tan\alpha.$$

Thus we deduce

$$\tan\alpha = \dfrac{1}{7}.$$

Next, consider the second relation

$$\sqrt{\dfrac{1-\cos 2\beta}{2}} = \dfrac{1}{\sqrt{10}}.$$

We recall the familiar identity

$$\sin^2\beta = \dfrac{1-\cos 2\beta}{2}.$$

Taking the positive square root (again justified by $$0<\beta<\dfrac{\pi}{2}$$) we get

$$\sin\beta = \dfrac{1}{\sqrt{10}}.$$

Using the Pythagorean relation $$\sin^2\beta + \cos^2\beta = 1,$$ we find

$$\cos\beta = \sqrt{1 - \sin^2\beta} = \sqrt{1 - \dfrac{1}{10}} = \sqrt{\dfrac{9}{10}} = \dfrac{3}{\sqrt{10}}.$$

Hence

$$\tan\beta = \dfrac{\sin\beta}{\cos\beta} = \dfrac{\dfrac{1}{\sqrt{10}}}{\dfrac{3}{\sqrt{10}}} = \dfrac{1}{3}.$$

Our goal is to compute $$\tan(\alpha + 2\beta).$$ We already know $$\tan\alpha$$ and $$\tan\beta,$$ so we next find $$\tan 2\beta.$$ The double-angle formula for tangent is

$$\tan 2\beta = \dfrac{2\tan\beta}{1-\tan^2\beta}.$$

Substituting $$\tan\beta = \dfrac{1}{3}$$ gives

$$\tan 2\beta = \dfrac{2\left(\dfrac{1}{3}\right)}{1-\left(\dfrac{1}{3}\right)^2} = \dfrac{\dfrac{2}{3}}{1-\dfrac{1}{9}} = \dfrac{\dfrac{2}{3}}{\dfrac{8}{9}} = \dfrac{2}{3}\times\dfrac{9}{8} = \dfrac{3}{4}.$$

Now we use the angle‐addition formula for tangent:

$$\tan(\alpha + 2\beta) = \dfrac{\tan\alpha + \tan 2\beta}{1 - \tan\alpha\,\tan 2\beta}.$$

We substitute $$\tan\alpha = \dfrac{1}{7}$$ and $$\tan 2\beta = \dfrac{3}{4}$$:

First the numerator:

$$\tan\alpha + \tan 2\beta = \dfrac{1}{7} + \dfrac{3}{4} = \dfrac{4}{28} + \dfrac{21}{28} = \dfrac{25}{28}.$$

Then the denominator:

$$1 - \tan\alpha\,\tan 2\beta = 1 - \left(\dfrac{1}{7}\right)\left(\dfrac{3}{4}\right) = 1 - \dfrac{3}{28} = \dfrac{25}{28}.$$

Hence

$$\tan(\alpha + 2\beta) = \dfrac{\dfrac{25}{28}}{\dfrac{25}{28}} = 1.$$

Hence, the correct answer is Option 1.

We are told that the three sides of a triangle have lengths $$5,\;5r,\;5r^{2}$$, where $$r\gt 0$$. For any three positive numbers to be the sides of a triangle, they must satisfy the triangle‐inequality theorem, which states:

Sum of any two sides > the remaining side.

Because the factor $$5$$ is common to all three expressions, it is convenient to divide each inequality by $$5$$. This leaves the simpler numbers $$1,\;r,\;r^{2}$$. Thus, the triangle inequalities become:

$$\begin{aligned} 1+r \gt r^{2},\\[4pt] r+r^{2} \gt 1,\\[4pt] 1+r^{2} \gt r. \end{aligned}$$

The third inequality $$1+r^{2}\gt r$$ is automatically true for every positive $$r,$$ so only the first two need detailed analysis. We examine them by separating the two natural cases $$r\ge 1$$ and $$0\lt r\le 1.$$

Case 1: $$r\ge 1.$$} In this range $$r^{2}\ge r\ge 1,$$ so $$r^{2}$$ is the largest of the three numbers $$1,\;r,\;r^{2}.$$ The pertinent inequality is therefore

$$1+r\gt r^{2}.$$

Rearranging gives

$$r^{2}-r-1\lt 0.$$

We solve the quadratic $$r^{2}-r-1=0$$ by the quadratic formula:

$$r=\frac{1\pm\sqrt{1+4}}{2}=\frac{1\pm\sqrt5}{2}.$$

Thus the two real roots are $$\dfrac{1-\sqrt5}{2}\;(\text{negative})$$ and $$\dfrac{1+\sqrt5}{2}\approx1.618.$$ A quadratic with positive leading coefficient is negative between its roots, so

$$-\,\frac{\sqrt5-1}{2}\lt r\lt \frac{1+\sqrt5}{2}.$$

Because we are in the subcase $$r\ge1,$$ the combined condition is

$$1\le r\lt \frac{1+\sqrt5}{2}\quad\Longrightarrow\quad1\le r\lt 1.618\ldots$$

Case 2: $$0\lt r\le1.$$} Here the largest of the three numbers $$1,\;r,\;r^{2}$$ is $$1.$$ Hence we must satisfy

$$r+r^{2}\gt 1.$$

Rewriting yields

$$r^{2}+r-1\gt 0.$$

We solve the quadratic $$r^{2}+r-1=0$$ in the same way:

$$r=\frac{-1\pm\sqrt{1+4}}{2}=\frac{-1\pm\sqrt5}{2}.$$

The positive root is $$\dfrac{\sqrt5-1}{2}\approx0.618.$$ Because the quadratic opens upward, it is positive outside the interval between its roots, giving

$$r\gt \frac{\sqrt5-1}{2}\quad\text{or}\quad r\lt -\frac{1+\sqrt5}{2}.$$

Only positive $$r$$ are relevant, so for this case

$$\frac{\sqrt5-1}{2}\lt r\le1\quad\Longrightarrow\quad0.618\ldots\lt r\le1.$$

Combining both cases we see that a valid $$r$$ must lie in the single continuous interval

$$0.618\ldots\lt r\lt 1.618\ldots$$

Now we test the four given choices:

$$\begin{aligned} \frac34 &=0.75 &\checkmark\\[4pt] \frac32 &=1.50 &\checkmark\\[4pt] \frac54 &=1.25 &\checkmark\\[4pt] \frac74 &=1.75 &\times \end{aligned}$$

The only option that lies outside the required range is $$\dfrac74.$$ Therefore $$r$$ cannot be equal to $$\dfrac74.$$ Hence, the correct answer is Option D.

We are given the equality

$$\sin^4\alpha + 4\cos^4\beta + 2 = 4\sqrt2\;\sin\alpha\cos\beta\,,\qquad \alpha,\,\beta\in[0,\pi].$$

Because $$\sin\alpha\ge 0$$ for every $$\alpha\in[0,\pi],$$ we set

$$\sin\alpha=\sqrt{s}\quad(s\ge 0),\qquad \cos\beta = C\quad(C\in[-1,1]).$$

Then $$s=\sin^2\alpha\in[0,1]$$ and $$C^2=\cos^2\beta\in[0,1].$$ Substituting these symbols in the given equation we obtain

$$s^2 + 4C^4 + 2 = 4\sqrt2\,\sqrt{s}\,C. \quad -(1)$$

The sign of the right-hand side is the sign of $$C$$. The left-hand side is clearly positive. Hence $$C$$ must be non-negative, i.e. $$C\ge 0\;(0\le\beta\le\pi/2).$$ Therefore we can write $$C=\sqrt{c}\,(c=C^2).$$ It is more convenient to remove the radicals completely, so we introduce two new non-negative variables

$$v=\sqrt{s}=\sin\alpha,\qquad u=\sqrt2\,C=\sqrt2\,\cos\beta.$$

Because $$0\le s\le 1$$ and $$0\le C\le 1,$$ we have $$0\le v\le 1,\;0\le u\le\sqrt2.$$ Now (1) becomes

$$v^4+u^4+2=4uv. \quad -(2)$$

To analyse (2) we separate two non-negative parts.

First write

$$v^4+u^4=(v^2-u^2)^2+2v^2u^2,$$ so that

$$(2)\;\Longrightarrow\;(v^2-u^2)^2+2v^2u^2+2-4uv=0. \quad -(3)$$

We now estimate the second bracket in (3) with the Arithmetic-Geometric Mean inequality. For any non-negative numbers $$x,y$$ we have $$\dfrac{x+1}{2}\ge\sqrt{x}.$$ Putting $$x=v^2u^2$$ yields

$$v^2u^2+1\;\ge\;2vu\quad\Longrightarrow\quad 2v^2u^2+2\;\ge\;4vu.$$

Hence

$$2v^2u^2+2-4vu\;\ge\;0.$$

Both summands in (3) are therefore non-negative, and their sum equals zero. Consequently each must vanish.

1. $$(v^2-u^2)^2=0\;\Longrightarrow\;v^2=u^2\;\Longrightarrow\;v=u.$$

2. $$2v^2u^2+2-4vu=0\;\Longrightarrow\;v^2u^2+1=2vu.$$ Substituting $$u=v$$ from step 1 we get $$v^4+1=2v^2\;\Longrightarrow\;(v^2-1)^2=0\;\Longrightarrow\;v=1.$$

Thus

$$v=1,\qquad u=1.$$ Reverting to the original trigonometric quantities, we have

$$\sin\alpha=v=1\;\Longrightarrow\;\alpha=\dfrac{\pi}{2},$$

$$\sqrt2\,\cos\beta=u=1\;\Longrightarrow\;\cos\beta=\dfrac1{\sqrt2}\; \Longrightarrow\;\beta=\dfrac{\pi}{4}.$$

Now we evaluate the required expression. First state the identity that will be used:

$$\cos(A+B)-\cos(A-B)=-2\sin A\sin B.$$

Using $$A=\alpha,\;B=\beta$$ together with $$\sin\alpha=1,\;\sin\beta=\dfrac1{\sqrt2},$$ we get

$$\cos(\alpha+\beta)-\cos(\alpha-\beta)= -2\sin\alpha\sin\beta =-2\left(1\right)\left(\dfrac1{\sqrt2}\right)=-\sqrt2.$$

Hence, the correct answer is Option B.

First, recall the well-known double-angle identity for sine:

$$\sin(2\theta)=2\sin\theta\cos\theta.$$

We shall apply this identity successively to break the sine of a larger angle into sines and cosines of smaller and smaller angles until the desired product appears.

Take $$\theta=\dfrac{\pi}{2^{2}}.$$ Then $$2\theta=\dfrac{\pi}{2},$$ and the identity gives

$$\sin\!\Bigl(\dfrac{\pi}{2}\Bigr)=2\sin\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr).$$

But $$\sin\!\bigl(\dfrac{\pi}{2}\bigr)=1,$$ so we obtain

$$1=2\sin\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr). \quad -(1)$$

Now apply the same identity once more, this time to the angle $$\dfrac{\pi}{2^{3}}.$$ Since $$2\!\left(\dfrac{\pi}{2^{3}}\right)=\dfrac{\pi}{2^{2}},$$ we get

$$\sin\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr)=2\sin\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr). \quad -(2)$$

Substituting the value of $$\sin\!\bigl(\dfrac{\pi}{2^{2}}\bigr)$$ from (2) into (1), we have

$$1=2\Bigl[2\sin\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr)\Bigr]\cos\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr) =2^{2}\sin\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr).$$

We can clearly see a pattern forming. Repeating this process again and again, each time halving the angle, produces additional cosine factors and an additional power of 2 in front. After proceeding down to the angle $$\dfrac{\pi}{2^{10}},$$ the cumulative result is

$$1 =2^{9}\sin\!\Bigl(\dfrac{\pi}{2^{10}}\Bigr)\, \cos\!\Bigl(\dfrac{\pi}{2^{10}}\Bigr) \cos\!\Bigl(\dfrac{\pi}{2^{9}}\Bigr) \cos\!\Bigl(\dfrac{\pi}{2^{8}}\Bigr) \cdots \cos\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr).$$

Writing the product of cosines in ascending order of the denominators and grouping conveniently, this is

$$1 =2^{9}\Bigl[\sin\!\Bigl(\dfrac{\pi}{2^{10}}\Bigr)\prod_{k=2}^{10}\cos\!\Bigl(\dfrac{\pi}{2^{k}}\Bigr)\Bigr].$$

Observe that the bracketed expression is exactly the quantity asked for in the question, namely

$$\cos\!\Bigl(\dfrac{\pi}{2^{2}}\Bigr)\cos\!\Bigl(\dfrac{\pi}{2^{3}}\Bigr)\cdots\cos\!\Bigl(\dfrac{\pi}{2^{10}}\Bigr)\sin\!\Bigl(\dfrac{\pi}{2^{10}}\Bigr).$$

Denote this required value by $$P.$$ From the equation just obtained we therefore have

$$1 = 2^{9} P \quad\Longrightarrow\quad P = \dfrac{1}{2^{9}} = \dfrac{1}{512}.$$

Hence, the correct answer is Option B.

We have to find the value of the expression $$\cos^2 10^\circ-\cos 10^\circ\cos 50^\circ+\cos^2 50^\circ.$$

First, we convert each squared cosine term by using the identity

$$\cos^2\theta=\frac{1+\cos 2\theta}{2}.$$

Applying this to the two squared terms gives

$$\cos^2 10^\circ=\frac{1+\cos 20^\circ}{2},\qquad \cos^2 50^\circ=\frac{1+\cos 100^\circ}{2}.$$

Substituting these into the original expression, we obtain

$$\cos^2 10^\circ-\cos 10^\circ\cos 50^\circ+\cos^2 50^\circ =\frac{1+\cos 20^\circ}{2}-\cos 10^\circ\cos 50^\circ +\frac{1+\cos 100^\circ}{2}.$$

Now we combine the two half-fractions:

$$\frac{1+\cos 20^\circ}{2}+\frac{1+\cos 100^\circ}{2} =\frac{2+\cos 20^\circ+\cos 100^\circ}{2} =1+\frac{\cos 20^\circ+\cos 100^\circ}{2}.$$

So the whole expression turns into

$$1+\frac{\cos 20^\circ+\cos 100^\circ}{2}-\cos 10^\circ\cos 50^\circ.$$

Next we simplify $$\cos 100^\circ$$. Since $$100^\circ=180^\circ-80^\circ,$$ we use $$\cos(180^\circ-\alpha)=-\cos\alpha$$ to get

$$\cos 100^\circ=-\cos 80^\circ.$$

Hence $$\cos 20^\circ+\cos 100^\circ=\cos 20^\circ-\cos 80^\circ,$$ and the expression becomes

$$1+\frac{\cos 20^\circ-\cos 80^\circ}{2}-\cos 10^\circ\cos 50^\circ.$$

To handle the product $$\cos 10^\circ\cos 50^\circ,$$ we employ the product-to-sum identity

$$\cos A\cos B=\frac{\cos(A+B)+\cos(A-B)}{2}.$$

Taking $$A=10^\circ, B=50^\circ$$ gives

$$\cos 10^\circ\cos 50^\circ =\frac{\cos(10^\circ+50^\circ)+\cos(10^\circ-50^\circ)}{2} =\frac{\cos 60^\circ+\cos(-40^\circ)}{2}.$$

Because $$\cos(-\alpha)=\cos\alpha,$$ we have $$\cos(-40^\circ)=\cos 40^\circ,$$ and since $$\cos 60^\circ=\frac12,$$ we get

$$\cos 10^\circ\cos 50^\circ =\frac{\frac12+\cos 40^\circ}{2} =\frac14+\frac{\cos 40^\circ}{2}.$$

Substituting this back, we arrive at

$$1+\frac{\cos 20^\circ-\cos 80^\circ}{2}-\left(\frac14+\frac{\cos 40^\circ}{2}\right) =1-\frac14+\frac{\cos 20^\circ-\cos 80^\circ-\cos 40^\circ}{2}.$$

The constant terms combine to give $$\frac34,$$ so we now focus on the bracket:

$$\cos 20^\circ-\cos 80^\circ-\cos 40^\circ.$$

We first simplify $$\cos 20^\circ-\cos 40^\circ$$ using the difference identity

$$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}.$$

Here $$A=20^\circ, B=40^\circ,$$ so

$$\cos 20^\circ-\cos 40^\circ =-2\sin\frac{20^\circ+40^\circ}{2}\sin\frac{20^\circ-40^\circ}{2} =-2\sin 30^\circ\sin(-10^\circ).$$

Because $$\sin 30^\circ=\frac12$$ and $$\sin(-10^\circ)=-\sin 10^\circ,$$ we get

$$\cos 20^\circ-\cos 40^\circ =-2\left(\frac12\right)(-\sin 10^\circ) =\sin 10^\circ.$$

So the whole bracket becomes

$$\sin 10^\circ-\cos 80^\circ.$$

But $$\cos 80^\circ=\sin 10^\circ$$ (since $$\cos\theta=\sin(90^\circ-\theta)$$), therefore

$$\sin 10^\circ-\cos 80^\circ=\sin 10^\circ-\sin 10^\circ=0.$$

Thus the entire numerator of the fraction is zero, and the fractional part vanishes. We are left with

$$\frac34+0=\frac34.$$

Hence, the correct answer is Option A.

We begin with the inequality

$$2^{\sqrt{\sin^2 x - 2\sin x + 5}}\;\cdot\;\frac{1}{4^{\sin^2 y}}\;\le\;1.$$

The denominator involves a power of $$4$$, and it is useful to rewrite it with the same base $$2$$. Since $$4 = 2^{2}$$, we have the identity

$$4^{\sin^2 y}\;=\;\bigl(2^2\bigr)^{\sin^2 y}\;=\;2^{2\sin^2 y}.$$

Substituting this into the given expression gives

$$2^{\sqrt{\sin^2 x - 2\sin x + 5}}\;\cdot\;2^{-2\sin^2 y}\;\le\;1,$$

which, by the basic rule $$a^m\cdot a^n = a^{m+n}$$, simplifies to

$$2^{\;\sqrt{\sin^2 x - 2\sin x + 5}\;-\;2\sin^2 y}\;\le\;1.$$

Now we recall the fact that for any base $$a \gt 1$$, the inequality

$$a^{k}\;\le\;1\quad\Longleftrightarrow\quad k\;\le\;0.$$

Using this fact with the base $$a=2$$, the exponent must satisfy

$$\sqrt{\sin^2 x - 2\sin x + 5}\;-\;2\sin^2 y\;\le\;0.$$

Thus

$$\sqrt{\sin^2 x - 2\sin x + 5}\;\le\;2\sin^2 y. \quad -(1)$$

We next simplify the quantity under the square-root. Completing the square gives

$$$ \sin^2 x \;-\;2\sin x \;+\;5 \;=\; \bigl(\sin x -1\bigr)^2 \;+\;4. $$$

Hence

$$\sqrt{\sin^2 x - 2\sin x + 5} \;=\; \sqrt{\bigl(\sin x -1\bigr)^2 + 4}.$$

The term $$(\sin x-1)^2$$ is always non-negative, so

$$$ (\sin x -1)^2 + 4 \;\ge\; 4, \quad\text{and therefore}\quad \sqrt{(\sin x -1)^2 + 4}\;\ge\;2. $$$

Returning to inequality (1), we know the left-hand side is at least $$2$$, so the right-hand side must also be at least $$2$$. Thus

$$2\sin^2 y\;\ge\;2 \;\;\Longrightarrow\;\; \sin^2 y\;\ge\;1.$$

But for every real number $$y$$ we have $$\sin^2 y \le 1$$. Combining $$\sin^2 y \ge 1$$ with $$\sin^2 y \le 1$$ forces

$$\sin^2 y = 1,$$ so $$\sin y = \pm 1.$$

With $$\sin^2 y = 1$$ we obtain $$2\sin^2 y = 2$$, and inequality (1) becomes

$$\sqrt{(\sin x -1)^2 + 4}\;\le\;2.$$

Since the square-root is never less than $$2$$, equality must occur:

$$\sqrt{(\sin x -1)^2 + 4} = 2.$$

Squaring both sides yields

$$(\sin x -1)^2 + 4 = 4 \quad\Longrightarrow\quad (\sin x -1)^2 = 0,$$

which simplifies to

$$\sin x - 1 = 0 \quad\Longrightarrow\quad \sin x = 1.$$

Hence every permissible pair $$(x,y)$$ satisfies

$$\sin x = 1,\qquad \sin y = \pm 1,$$ which in turn gives

$$|\sin x| = 1 = |\sin y|.$$

Therefore the common relation that all such pairs obey is

$$|\sin x| = |\sin y|,$$ which is exactly the statement in Option C.

Hence, the correct answer is Option C.

We have the family of functions $$f_k(x)=\dfrac{1}{k}\bigl(\sin^{\,k}x+\cos^{\,k}x\bigr)$$ for positive integers $$k$$. We need the expression $$f_4(x)-f_6(x)$$ for every real $$x$$.

First we write the two required members of the family explicitly:

$$f_4(x)=\dfrac{1}{4}\bigl(\sin^{4}x+\cos^{4}x\bigr),\qquad f_6(x)=\dfrac{1}{6}\bigl(\sin^{6}x+\cos^{6}x\bigr).$$

To simplify $$\sin^{4}x+\cos^{4}x$$, we start from the square of the Pythagorean identity. Since $$\sin^{2}x+\cos^{2}x=1$$, squaring both sides gives

$$\bigl(\sin^{2}x+\cos^{2}x\bigr)^{2}=1^{2}=1.$$

Expanding the left side,

$$\sin^{4}x+2\sin^{2}x\cos^{2}x+\cos^{4}x=1.$$

Rearranging,

$$\sin^{4}x+\cos^{4}x=1-2\sin^{2}x\cos^{2}x.$$

Next we simplify $$\sin^{6}x+\cos^{6}x$$. We use the algebraic identity for the sum of cubes, namely $$a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$$. Let

$$a=\sin^{2}x,\quad b=\cos^{2}x.$$

Then

$$\sin^{6}x+\cos^{6}x=(\sin^{2}x)^{3}+(\cos^{2}x)^{3} =(a+b)\bigl(a^{2}-ab+b^{2}\bigr).$$

We already know $$a+b=\sin^{2}x+\cos^{2}x=1$$, so only $$a^{2}-ab+b^{2}$$ remains. Observe

$$a^{2}-ab+b^{2}=\sin^{4}x-\sin^{2}x\cos^{2}x+\cos^{4}x.$$

Substituting the earlier result $$\sin^{4}x+\cos^{4}x=1-2\sin^{2}x\cos^{2}x$$ gives

$$a^{2}-ab+b^{2}=(1-2\sin^{2}x\cos^{2}x)-\sin^{2}x\cos^{2}x =1-3\sin^{2}x\cos^{2}x.$$

Hence

$$\sin^{6}x+\cos^{6}x=(a+b)\bigl(a^{2}-ab+b^{2}\bigr) =1\bigl(1-3\sin^{2}x\cos^{2}x\bigr)=1-3\sin^{2}x\cos^{2}x.$$

For brevity, let us denote $$\sin^{2}x\cos^{2}x$$ by $$t$$. Therefore

$$\sin^{4}x+\cos^{4}x=1-2t,\qquad \sin^{6}x+\cos^{6}x=1-3t.$$

We now substitute these into $$f_4(x)$$ and $$f_6(x)$$:

$$f_4(x)=\dfrac{1}{4}(1-2t),\qquad f_6(x)=\dfrac{1}{6}(1-3t).$$

The desired difference becomes

$$f_4(x)-f_6(x)=\dfrac{1}{4}(1-2t)-\dfrac{1}{6}(1-3t).$$

To combine the fractions, we take the common denominator $$12$$:

$$f_4(x)-f_6(x)=\dfrac{3}{12}(1-2t)-\dfrac{2}{12}(1-3t).$$

Expanding the numerators,

$$=\dfrac{3(1-2t)-2(1-3t)}{12} =\dfrac{3-6t-2+6t}{12}.$$

The $$-6t$$ and $$+6t$$ cancel out, leaving

$$=\dfrac{1}{12}.$$

This final expression contains no $$x$$, confirming that the difference is the same for all real $$x$$. Hence, the correct answer is Option A.

We begin with the given equation

$$\cos 2x + \alpha \sin x \;=\; 2\alpha - 7.$$

First, we rewrite $$\cos 2x$$ in terms of $$\sin x$$ by using the standard identity

$$\cos 2x \;=\; 1 - 2\sin^{2}x.$$

Substituting this identity, we obtain

$$1 - 2\sin^{2}x + \alpha \sin x \;=\; 2\alpha - 7.$$

For simplicity, let us set

$$y \;=\; \sin x,$$

remembering that $$y$$ must always satisfy $$-1 \le y \le 1.$$ With this substitution the equation becomes

$$1 - 2y^{2} + \alpha y \;=\; 2\alpha - 7.$$

We bring every term to the left-hand side so that the expression equals zero:

$$1 - 2y^{2} + \alpha y - 2\alpha + 7 \;=\; 0.$$

Combining the constant terms $$1 + 7 = 8,$$ we have

$$-2y^{2} + \alpha y + 8 - 2\alpha = 0.$$

Multiplying by $$-1$$ to make the coefficient of $$y^{2}$$ positive, we get

$$2y^{2} - \alpha y - 8 + 2\alpha = 0,$$

or, more neatly,

$$2y^{2} - \alpha y + 2\alpha - 8 = 0.$$

This is a quadratic equation in $$y$$. We want some $$y$$ in the interval $$[-1,\,1]$$ to satisfy it. To find all such $$\alpha,$$ we solve the quadratic for $$\alpha$$ in terms of $$y$$ (treating $$y$$ as the variable and $$\alpha$$ as the unknown constant). Collecting the $$\alpha$$ terms, we write

$$- \alpha y + 2\alpha = \alpha(2 - y).$$

Shifting the remaining terms to the other side gives

$$\alpha(2 - y) = 8 - 2y^{2}.$$

Provided $$y \ne 2$$ (and indeed $$y$$ is never $$2$$ because $$-1 \le y \le 1$$), we can divide to obtain

$$\displaystyle \alpha = \frac{8 - 2y^{2}}{\,2 - y\,}.$$

Factoring a $$2$$ from the numerator yields the simpler expression

$$\alpha = 2\,\frac{4 - y^{2}}{2 - y}.$$\p>

Define the function

$$g(y) = 2\,\frac{4 - y^{2}}{2 - y}, \qquad -1 \le y \le 1.$$

The set $$S$$ that we seek is precisely the set of all values taken by $$g(y)$$ as $$y$$ varies over $$[-1,\,1].$$ To determine this set we examine the monotonicity of $$g(y).$$ Write

$$g(y) = 2\,\frac{N(y)}{D(y)},$$

where $$N(y) = 4 - y^{2}$$ and $$D(y) = 2 - y.$$ Using the quotient rule, the derivative of the inner fraction is

$$\frac{d}{dy}\!\left(\frac{N(y)}{D(y)}\right) = \frac{N'(y)D(y) - N(y)D'(y)}{D(y)^{2}}.$$

We compute each part:

$$N'(y) = -2y, \qquad D'(y) = -1.$$

Substituting these, we obtain

$$\frac{d}{dy}\!\left(\frac{N}{D}\right) = \frac{(-2y)(2 - y) - (4 - y^{2})(-1)}{(2 - y)^{2}} = \frac{-4y + 2y^{2} + 4 - y^{2}}{(2 - y)^{2}} = \frac{y^{2} - 4y + 4}{(2 - y)^{2}} = \frac{(y - 2)^{2}}{(2 - y)^{2}}.$$

Since both numerator and denominator are perfect squares, their ratio is always non-negative and equals zero only when $$y = 2,$$ which is outside the interval $$[-1,\,1].$$ Hence the derivative is strictly positive on $$[-1,\,1]$$ and $$g(y)$$ is strictly increasing on that interval.

Because $$g$$ is increasing, its minimum and maximum values on $$[-1,\,1]$$ occur at the endpoints:

For $$y = -1:$$

$$g(-1) = 2\,\frac{4 - (-1)^{2}}{2 - (-1)} = 2\,\frac{4 - 1}{3} = 2 \times \frac{3}{3} = 2.$$

For $$y = 1:$$

$$g(1) = 2\,\frac{4 - 1^{2}}{2 - 1} = 2\,\frac{3}{1} = 6.$$

Because every intermediate value is attained by continuity and monotonicity, the range of $$g$$ is the closed interval

$$[2,\,6].$$

Therefore the set $$S$$ of all real numbers $$\alpha$$ for which the original trigonometric equation admits a solution is

$$S = [2,\,6].$$

Hence, the correct answer is Option B.

We are asked to find every $$\theta$$ lying in the closed interval $$[-2\pi,\,2\pi]$$ that satisfies the trigonometric equation

$$2\cos^2\theta + 3\sin\theta = 0.$$

First, we convert the entire equation into a single trigonometric function. We recall the Pythagorean identity

$$\sin^2\theta + \cos^2\theta = 1 \quad\Longrightarrow\quad \cos^2\theta = 1 - \sin^2\theta.$$

Substituting $$\cos^2\theta = 1 - \sin^2\theta$$ in the given equation, we obtain

$$2(1 - \sin^2\theta) + 3\sin\theta = 0.$$

Simplifying term by term, we have

$$2 - 2\sin^2\theta + 3\sin\theta = 0.$$

Let us set $$x = \sin\theta$$ for convenience. Rewriting the equation in terms of $$x$$ gives

$$2 - 2x^2 + 3x = 0.$$

Next, we collect like terms and arrange them in descending powers of $$x$$:

$${-2x^2} + 3x + 2 = 0.$$

Multiplying every term by $$-1$$ (which does not change the solution set) yields the more familiar quadratic form

$$2x^2 - 3x - 2 = 0.$$

We solve this quadratic equation with the quadratic formula. For $$ax^2 + bx + c = 0,$$ the roots are

$$x = \frac{-b \pm \sqrt{\,b^2 - 4ac\,}}{2a}.$$

Here $$a = 2,\; b = -3,\; c = -2.$$ Substituting these values, we find

$$x = \frac{-(-3) \pm \sqrt{\,(-3)^2 - 4(2)(-2)\,}}{2(2)} = \frac{3 \pm \sqrt{\,9 + 16\,}}{4} = \frac{3 \pm \sqrt{25}}{4} = \frac{3 \pm 5}{4}.$$

This gives two possible roots:

$$x_1 = \frac{3 + 5}{4} = \frac{8}{4} = 2, \qquad x_2 = \frac{3 - 5}{4} = \frac{-2}{4} = -\frac12.$$

Since $$x = \sin\theta$$ must lie in the interval $$[-1,\,1],$$ the value $$x = 2$$ is impossible and must be discarded. Thus the only admissible value is

$$\sin\theta = -\frac12.$$

We now solve $$\sin\theta = -\dfrac12$$ for all $$\theta$$ in the range $$[-2\pi,\,2\pi].$$ On the unit circle, the sine function equals $$-\dfrac12$$ at the following standard positions:

$$\theta = -\frac\pi6 + 2k\pi, \qquad\theta = -\frac{5\pi}6 + 2k\pi,$$

where $$k \in \mathbb Z.$$ These are the fourth- and third-quadrant angles corresponding to sine $$-\dfrac12.$$ We list integral values of $$k$$ that keep $$\theta$$ within $$[-2\pi,\,2\pi]$$ and record each distinct angle.

Taking $$k = 0$$

$$\theta_1 = -\frac\pi6, \qquad \theta_2 = -\frac{5\pi}6.$$

Taking $$k = 1$$

$$\theta_3 = -\frac\pi6 + 2\pi = -\frac\pi6 + \frac{12\pi}6 = \frac{11\pi}6,$$

$$\theta_4 = -\frac{5\pi}6 + 2\pi = -\frac{5\pi}6 + \frac{12\pi}6 = \frac{7\pi}6.$$

Taking $$k = -1$$ would produce

$$\theta = -\frac\pi6 - 2\pi = -\frac{13\pi}6 \lt -2\pi,$$

$$\theta = -\frac{5\pi}6 - 2\pi = -\frac{17\pi}6 \lt -2\pi,$$

both of which lie outside the prescribed interval, so they are excluded. Similarly, $$k = 2$$ gives angles greater than $$2\pi.$$ Therefore the complete solution set is

$$S = \Bigl\{-\frac{5\pi}6,\; -\frac\pi6,\; \frac{7\pi}6,\; \frac{11\pi}6\Bigr\}.$$

We now compute the sum of these four angles:

$$\Bigl(-\frac{5\pi}6\Bigr) + \Bigl(-\frac\pi6\Bigr) + \frac{7\pi}6 + \frac{11\pi}6 = \left(-\frac{6\pi}6\right) + \frac{18\pi}6 = -\pi + 3\pi = 2\pi.$$

Thus the sum of all elements of $$S$$ equals $$2\pi.$$ Among the given options, this value corresponds to Option D.

Hence, the correct answer is Option D.

We need all angles $$\theta$$ lying in the open interval $$(0,\dfrac{\pi}{2})$$ that satisfy the trigonometric equation

$$\sin^2 2\theta+\cos^4 2\theta=\dfrac34.$$

First, to simplify the expression, we put

$$x=2\theta.$$

Because $$\theta\in(0,\dfrac{\pi}{2})$$, doubling every number in this interval gives

$$x\in(0,\pi).$$

Under this substitution the equation becomes

$$\sin^2 x+\cos^4 x=\dfrac34.$$

Next, we use the Pythagorean identity $$\sin^2 x+\cos^2 x=1.$$ Solving this for $$\sin^2 x$$ gives

$$\sin^2 x=1-\cos^2 x.$$

Substituting this in the transformed equation produces

$$\bigl(1-\cos^2 x\bigr)+\cos^4 x=\dfrac34.$$

Now let us introduce another temporary variable to avoid writing $$\cos$$ repeatedly. Put

$$y=\cos^2 x.$$

Because $$x\in(0,\pi),$$ the value of $$\cos x$$ ranges from ±1 down to 0, so $$y=\cos^2 x$$ must satisfy $$0<y<1.$$

Under this change of variable, the equation simplifies to a quadratic:

$$1-y+y^2=\dfrac34.$$

We move every term to the left so that the right-hand side becomes zero:

$$1-y+y^2-\dfrac34=0.$$

Combining the constant terms on the left, we have

$$y^2-y+\bigl(1-\dfrac34\bigr)=0,$$

so

$$y^2-y+\dfrac14=0.$$

Observe that the left side is a perfect square. Indeed,

$$(y-\dfrac12)^2=y^2-y+\dfrac14.$$

Therefore we can rewrite the quadratic as

$$(y-\dfrac12)^2=0.$$

Solving this gives a single real root

$$y=\dfrac12.$$

But $$y=\cos^2 x,$$ so

$$\cos^2 x=\dfrac12.$$

Taking square roots on both sides, we obtain

$$\cos x=\pm\dfrac1{\sqrt2}.$$

Now we must find every $$x$$ in the interval $$(0,\pi)$$ that satisfies this relation.

We know from basic trigonometry that

$$\cos x=\dfrac1{\sqrt2}\quad\text{at}\quad x=\dfrac{\pi}{4},$$

and

$$\cos x=-\dfrac1{\sqrt2}\quad\text{at}\quad x=\dfrac{3\pi}{4}.$$

Both of these angles indeed lie strictly between $$0$$ and $$\pi,$$ so the corresponding values of $$x$$ are

$$x_1=\dfrac{\pi}{4},\qquad x_2=\dfrac{3\pi}{4}.$$

We now reverse the substitution $$x=2\theta$$ to back-substitute for $$\theta$$:

For $$x_1$$:

$$2\theta_1=\dfrac{\pi}{4}\;\Longrightarrow\;\theta_1=\dfrac{\pi}{8}.$$

For $$x_2$$:

$$2\theta_2=\dfrac{3\pi}{4}\;\Longrightarrow\;\theta_2=\dfrac{3\pi}{8}.$$

Both values $$\dfrac{\pi}{8}$$ and $$\dfrac{3\pi}{8}$$ lie inside the required interval $$(0,\dfrac{\pi}{2}),$$ so they are the only solutions for $$\theta.$$

The question asks for the sum of all such $$\theta.$$ Adding the two values we get

$$\theta_1+\theta_2=\dfrac{\pi}{8}+\dfrac{3\pi}{8}=\dfrac{4\pi}{8}=\dfrac{\pi}{2}.$$

Hence, the correct answer is Option A.

We have to find the numerical value of the product $$\sin 10^\circ\,\sin 30^\circ\,\sin 50^\circ\,\sin 70^\circ.$$

First, notice that the three angles $$10^\circ,\;50^\circ,\;70^\circ$$ can be written in the symmetric form $$x,\;60^\circ-x,\;60^\circ+x$$ if we choose $$x = 10^\circ.$$ This observation suggests using a standard identity for the product $$\sin x\;\sin(60^\circ-x)\;\sin(60^\circ+x).$$

To derive that identity, we begin with the product-to-sum formula

$$\sin A\;\sin B = \tfrac12\bigl[\cos(A-B)-\cos(A+B)\bigr].$$

Put $$A = 60^\circ+x,\;B = 60^\circ-x.$$ Then

$$\sin(60^\circ+x)\,\sin(60^\circ-x) = \tfrac12\Bigl[\cos\bigl((60^\circ+x)-(60^\circ-x)\bigr) -\cos\bigl((60^\circ+x)+(60^\circ-x)\bigr)\Bigr].$$

Simplifying the arguments of the cosines, we get

$$\cos\bigl((60^\circ+x)-(60^\circ-x)\bigr)=\cos(2x),$$ $$\cos\bigl((60^\circ+x)+(60^\circ-x)\bigr)=\cos(120^\circ).$$

Because $$\cos120^\circ=-\tfrac12,$$ the product becomes

$$\sin(60^\circ+x)\,\sin(60^\circ-x) = \tfrac12\bigl[\cos 2x - (-\tfrac12)\bigr] = \tfrac12\cos 2x + \tfrac14.$$

Multiplying this result by $$\sin x,$$ we obtain

$$\sin x\,\sin(60^\circ-x)\,\sin(60^\circ+x) = \sin x\bigl(\tfrac12\cos 2x + \tfrac14\bigr).$$

Next we rewrite $$\cos 2x$$ via the double-angle identity $$\cos 2x = 1 - 2\sin^2 x.$$ Substituting,

$$\tfrac12\cos 2x + \tfrac14 = \tfrac12(1 - 2\sin^2 x) + \tfrac14 = \tfrac12 - \sin^2 x + \tfrac14 = \tfrac34 - \sin^2 x.$$

Hence

$$\sin x\,\sin(60^\circ-x)\,\sin(60^\circ+x) = \sin x\bigl(\tfrac34 - \sin^2 x\bigr) = \tfrac34\sin x - \sin^3 x.$$

Now we recall the triple-angle formula

$$\sin 3x = 3\sin x - 4\sin^3 x.$$

Dividing both sides of this formula by 4 gives

$$\tfrac14\sin 3x = \tfrac34\sin x - \sin^3 x.$$

Comparing with the expression just obtained, we see that

$$\boxed{\;\sin x\,\sin(60^\circ-x)\,\sin(60^\circ+x) = \tfrac14\sin 3x\;}.$$

Now set $$x = 10^\circ.$$ Then $$60^\circ-x = 50^\circ$$ and $$60^\circ+x = 70^\circ.$$ Therefore

$$\sin10^\circ\,\sin50^\circ\,\sin70^\circ = \tfrac14\sin 3(10^\circ) = \tfrac14\sin30^\circ.$$

Because $$\sin30^\circ = \tfrac12,$$ the triple product equals

$$\sin10^\circ\,\sin50^\circ\,\sin70^\circ = \tfrac14 \times \tfrac12 = \tfrac18.$$

The given expression contains one more factor $$\sin30^\circ=\tfrac12.$$ Multiplying this with the above result, we find

$$\sin10^\circ\,\sin30^\circ\,\sin50^\circ\,\sin70^\circ = \bigl(\tfrac18\bigr)\times\bigl(\tfrac12\bigr) = \tfrac1{16}.$$

Hence, the correct answer is Option B.

We need to find the number of values of $$x$$ in $$[0, \frac{\pi}{2})$$ satisfying $$\sin x - \sin 2x + \sin 3x = 0$$.

We use the sum-to-product identity. Group the first and third terms:

$$\sin x + \sin 3x = 2 \sin\left(\frac{x + 3x}{2}\right) \cos\left(\frac{3x - x}{2}\right) = 2 \sin 2x \cos x$$

Substituting back into the equation:

$$2 \sin 2x \cos x - \sin 2x = 0$$

Factoring out $$\sin 2x$$:

$$\sin 2x (2\cos x - 1) = 0$$

This gives us two cases:

Case 1: $$\sin 2x = 0$$

$$2x = n\pi$$ where $$n$$ is an integer, so $$x = \frac{n\pi}{2}$$.

In the interval $$[0, \frac{\pi}{2})$$: $$x = 0$$ (when $$n = 0$$) is valid. Note that $$x = \frac{\pi}{2}$$ (when $$n = 1$$) is NOT included since the domain is strictly less than $$\frac{\pi}{2}$$.

Case 2: $$2\cos x - 1 = 0$$