JEE Inverse Trigonometric Functions Questions

$$\alpha = \sin^{-1}\frac{3}{5}$$ $$\beta = \sin^{-1}\frac{5}{13}$$ $$\gamma = \sin^{-1}\frac{33}{65}$$

From these, we can find the sine and cosine values for each angle since they lie in the first quadrant:

$$\sin\alpha = \frac{3}{5}, \quad \cos\alpha = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}$$ $$\sin\beta = \frac{5}{13}, \quad \cos\beta = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}$$ $$\sin\gamma = \frac{33}{65}, \quad \cos\gamma = \sqrt{1 - \left(\frac{33}{65}\right)^2} = \frac{56}{65}$$

Next, we can calculate $$\sin(\alpha + \beta)$$ and $$\cos(\alpha + \beta)$$ using the trigonometric sum formulas:

$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \left(\frac{3}{5}\right)\left(\frac{12}{13}\right) + \left(\frac{4}{5}\right)\left(\frac{5}{13}\right) = \frac{36 + 20}{65} = \frac{56}{65}$$ $$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) = \frac{48 - 15}{65} = \frac{33}{65}$$

Now, we rewrite the original expression as $$\cos((\alpha + \beta) + \gamma)$$ and expand it:

$$\cos(\alpha + \beta + \gamma) = \cos(\alpha + \beta)\cos\gamma - \sin(\alpha + \beta)\sin\gamma$$

Substitute the known values into the equation:

$$\cos(\alpha + \beta + \gamma) = \left(\frac{33}{65}\right)\left(\frac{56}{65}\right) - \left(\frac{56}{65}\right)\left(\frac{33}{65}\right) = 0$$

Thus, the value of the expression is:$$0$$

 For all $$|x| \geq 1$$, we have:

$$\sec^{-1}x + \csc^{-1}x = \frac{\pi}{2}$$

Setting up variables: Let $$\alpha = \sec^{-1}x$$. Then $$\csc^{-1}x = \frac{\pi}{2} - \alpha$$.

Range of $$\alpha$$: The principal value of $$\sec^{-1}x$$ lies in $$[0, \pi] \setminus \{\frac{\pi}{2}\}$$.

For $$x \geq 1$$: $$\alpha \in \left[0, \frac{\pi}{2}\right)$$

For $$x \leq -1$$: $$\alpha \in \left(\frac{\pi}{2}, \pi\right]$$

So the full domain is $$\alpha \in \left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$$.

Expressing the function:

$$f(\alpha) = 16\left[\alpha^2 + \left(\frac{\pi}{2} - \alpha\right)^2\right]$$

$$= 16\left[\alpha^2 + \frac{\pi^2}{4} - \pi\alpha + \alpha^2\right]$$

$$= 16\left[2\alpha^2 - \pi\alpha + \frac{\pi^2}{4}\right]$$

$$= 32\alpha^2 - 16\pi\alpha + 4\pi^2$$

Finding the critical point:

$$f'(\alpha) = 64\alpha - 16\pi = 0$$

$$\alpha = \frac{\pi}{4}$$

Since $$f''(\alpha) = 64 > 0$$, this is a minimum. The point $$\alpha = \frac{\pi}{4}$$ lies in the domain (it is in $$[0, \frac{\pi}{2})$$).

Minimum value:$$f\left(\frac{\pi}{4}\right) = 32 \cdot \frac{\pi^2}{16} - 16\pi \cdot \frac{\pi}{4} + 4\pi^2$$

$$= 2\pi^2 - 4\pi^2 + 4\pi^2 = 2\pi^2$$

Maximum value: Since $$f(\alpha)$$ is a parabola opening upward with vertex at $$\alpha = \frac{\pi}{4}$$, the maximum on the domain occurs at the endpoint farthest from $$\frac{\pi}{4}$$.

At $$\alpha = 0$$: $$f(0) = 0 - 0 + 4\pi^2 = 4\pi^2$$

At $$\alpha = \pi$$: $$f(\pi) = 32\pi^2 - 16\pi^2 + 4\pi^2 = 20\pi^2$$

The maximum value is $$20\pi^2$$ (at $$\alpha = \pi$$, i.e., $$x = -1$$).

Sum of maximum and minimum $$= 20\pi^2 + 2\pi^2 = 22\pi^2$$

This matches Option B.

Write the numerical coefficients in the given expression with their trigonometric values:
$$\frac{\sqrt3}{2} = \cos\frac{\pi}{6}, \qquad \frac12 = \sin\frac{\pi}{6}$$

Let $$\alpha = \sin^{-1}x$$. By definition of the principal branch,
$$\alpha \in \left[-\frac{\pi}{2},\,\frac{\pi}{2}\right]$$ and $$\sin\alpha = x$$.

Because $$\alpha$$ lies between $$-\dfrac{\pi}{2}$$ and $$\dfrac{\pi}{2}$$, its cosine is non-negative, hence
$$\cos\alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1 - x^2}.$$\

Rewrite the inside of $$\sin^{-1}(\; \cdot \;)$$ using $$\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$$:

$$\frac{\sqrt3}{2}\,x + \frac12\sqrt{1 - x^2} \;=\; \sin\alpha\cos\frac{\pi}{6} + \cos\alpha\sin\frac{\pi}{6} \;=\; \sin\!\left(\alpha + \frac{\pi}{6}\right).$$

Therefore
$$\sin^{-1}\!\left(\frac{\sqrt3}{2}\,x + \frac12\sqrt{1 - x^2}\right) = \sin^{-1}\!\left(\sin\!\bigl(\alpha + \tfrac{\pi}{6}\bigr)\right).$$

We must now check whether $$\alpha + \dfrac{\pi}{6}$$ itself lies in the principal range $$\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right]$$ so that $$\sin^{-1}(\sin\theta)=\theta$$ holds unaltered.

Given $$-\dfrac12 \lt x \lt \dfrac1{\sqrt2}\;,$$ we have
$$-\frac{\pi}{6} \lt \alpha = \sin^{-1}x \lt \frac{\pi}{4}.$$

Add $$\dfrac{\pi}{6}$$ throughout:
$$0 \lt \alpha + \frac{\pi}{6} \lt \frac{\pi}{4} + \frac{\pi}{6} = \frac{5\pi}{12}.$$

The interval $$\left(0,\dfrac{5\pi}{12}\right)$$ is completely within the principal interval $$\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right]$$. Hence

$$\sin^{-1}\!\left(\sin\!\bigl(\alpha + \tfrac{\pi}{6}\bigr)\right) = \alpha + \frac{\pi}{6}.$$

Finally substitute back $$\alpha = \sin^{-1}x$$:

$$\sin^{-1}\!\left(\frac{\sqrt3}{2}\,x + \frac12\sqrt{1 - x^2}\right) = \frac{\pi}{6} + \sin^{-1}x.$$

Thus the expression equals $$\dfrac{\pi}{6} + \sin^{-1}x$$, which matches Option B.

The given integral is$$\alpha=\int_{1/2}^{2}\frac{\tan^{-1}x}{2x^{2}-3x+2}\,dx.$$

Step 1 : Use the reciprocal-substitution symmetry.
Put $$x=\frac{1}{t}$$ so that $$dx=-\frac{dt}{t^{2}}.$$ The limits change as $$x : 1/2 \to 2$$ implies $$t : 2 \to 1/2$$.

Then

$$\alpha=\int_{2}^{1/2}\frac{\tan^{-1}(1/t)}{2/t^{2}-3/t+2}\left(-\frac{dt}{t^{2}}\right) =\int_{1/2}^{2}\frac{\tan^{-1}(1/x)}{2x^{2}-3x+2}\,dx.$$

Step 2 : Add the two forms of the integral.

Adding the original and the transformed integrals,

$$2\alpha=\int_{1/2}^{2}\frac{\tan^{-1}x+\tan^{-1}(1/x)}{2x^{2}-3x+2}\,dx.$$

For every positive $$x\neq1$$, $$\tan^{-1}x+\tan^{-1}\!\left(\frac1x\right)=\frac{\pi}{2}.$$ (The value at $$x=1$$ does not affect the integral.)

Hence

$$2\alpha=\frac{\pi}{2}\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2},\qquad \alpha=\frac{\pi}{4}\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2}.$$

Step 3 : Evaluate the remaining integral.

Rewrite the quadratic:

$$2x^{2}-3x+2=2\!\left[x^{2}-\frac{3}{2}x+1\right] =2\!\Bigl[(x-\tfrac34)^{2}+\tfrac7{16}\Bigr] =2\!\Bigl[(x-\tfrac34)^{2}+(\tfrac{\sqrt7}{4})^{2}\Bigr].$$

Therefore

$$\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2} =\frac12\int_{1/2}^{2}\frac{dx}{(x-\tfrac34)^{2}+(\tfrac{\sqrt7}{4})^{2}} =\frac12\cdot\frac4{\sqrt7}\Bigl[\tan^{-1}\!\bigl(\tfrac{4x-3}{\sqrt7}\bigr)\Bigr]_{1/2}^{2}.$$

Simplifying the constant factor gives $$\dfrac{2}{\sqrt7}$$. Now compute the bracketed values:

At $$x=2$$: $$\dfrac{4x-3}{\sqrt7}=\dfrac{5}{\sqrt7}$$.
At $$x=\tfrac12$$: $$\dfrac{4x-3}{\sqrt7}=-\dfrac1{\sqrt7}$$.

Hence

$$\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2} =\frac{2}{\sqrt7}\Bigl[\tan^{-1}\!\bigl(\tfrac{5}{\sqrt7}\bigr)+\tan^{-1}\!\bigl(\tfrac{1}{\sqrt7}\bigr)\Bigr].$$

Using the addition formula $$\tan^{-1}a+\tan^{-1}b=\tan^{-1}\!\bigl(\dfrac{a+b}{1-ab}\bigr)$$ (since $$ab\lt 1$$),

$$\tan^{-1}\!\bigl(\tfrac{5}{\sqrt7}\bigr)+\tan^{-1}\!\bigl(\tfrac{1}{\sqrt7}\bigr) =\tan^{-1}\!\Bigl(\frac{\,\tfrac{5}{\sqrt7}+\tfrac{1}{\sqrt7}}{1-\tfrac{5}{7}}\Bigr) =\tan^{-1}\!\Bigl(\frac{\,\tfrac{6}{\sqrt7}}{\tfrac{2}{7}}\Bigr) =\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr).$$

Thus

$$\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2} =\frac{2}{\sqrt7}\,\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr).$$

Step 4 : Obtain $$\alpha$$.

$$\alpha=\frac{\pi}{4}\times\frac{2}{\sqrt7}\, \tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr) =\frac{\pi}{2\sqrt7}\, \tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr).$$

Step 5 : Evaluate the required expression.

Compute the angle inside the tangent:

$$\frac{2\alpha\sqrt7}{\pi} =\frac{2}{\pi}\cdot\frac{\pi}{2\sqrt7}\,\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr)\!\cdot\sqrt7 =\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr).$$

Therefore

$$\sqrt7\,\tan\!\Bigl(\frac{2\alpha\sqrt7}{\pi}\Bigr) =\sqrt7\,\tan\!\left[\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr)\right] =\sqrt7\cdot\frac{21}{\sqrt7}=21.$$

Hence, the required value is 21.

Write the required value as

$$\cot^{-1}\!\left(\dfrac{\sqrt{1+\tan^{2}2}-1}{\tan 2}\right)\;-\;\cot^{-1}\!\left(\dfrac{\sqrt{1+\tan^{2}\!\left(\dfrac12\right)}+1}{\tan\dfrac12}\right)$$

The standard identity is $$\sqrt{1+\tan^{2}\theta}=|\sec \theta|$$. We must keep the modulus because $$\sec\theta$$ can be negative although the square root is always positive.

Case 1: $$\theta = 2 \text{ rad}$$ (first inverse-cot term)

Because $$\cos2\lt0$$, $$|\sec2|=-\sec2$$. Hence

$$\dfrac{\sqrt{1+\tan^{2}2}-1}{\tan2}= \dfrac{-\sec2-1}{\tan2}= -\dfrac{\sec2+1}{\tan2}$$

Now use the algebraic simplification

$$\dfrac{\sec x+1}{\tan x}= \dfrac{\dfrac1{\cos x}+1}{\dfrac{\sin x}{\cos x}}= \dfrac{1+\cos x}{\sin x}= \dfrac{2\cos^{2}\dfrac x2}{2\sin\dfrac x2\cos\dfrac x2}= \cot\dfrac x2$$

Putting $$x=2$$ gives $$\dfrac{\sec2+1}{\tan2}=\cot1$$, so

$$\dfrac{\sqrt{1+\tan^{2}2}-1}{\tan2}= -\cot1 = \cot\!\bigl(\pi-1\bigr)$$

(since $$\cot(\pi-\alpha)=-\cot\alpha$$). The principal branch of $$\cot^{-1}y$$ lies in $$(0,\pi)$$ and is monotonic, therefore

$$\cot^{-1}\!\left(\dfrac{\sqrt{1+\tan^{2}2}-1}{\tan2}\right)=\cot^{-1}\!\bigl(\cot(\pi-1)\bigr)=\pi-1$$

Case 2: $$\theta=\dfrac12 \text{ rad}$$ (second inverse-cot term)

Here $$\cos\dfrac12\gt0$$, so $$|\sec\dfrac12|=\sec\dfrac12$$. Thus

$$\dfrac{\sqrt{1+\tan^{2}\dfrac12}+1}{\tan\dfrac12}= \dfrac{\sec\dfrac12+1}{\tan\dfrac12}=\cot\dfrac14$$

Because $$0\lt \dfrac14\lt \dfrac\pi2$$, the principal value is simply

$$\cot^{-1}\!\left(\dfrac{\sqrt{1+\tan^{2}\dfrac12}+1}{\tan\dfrac12}\right)=\cot^{-1}\!\bigl(\cot\dfrac14\bigr)=\dfrac14$$

Final value

$$\bigl(\pi-1\bigr)-\dfrac14=\pi-\dfrac54$$

Therefore the required value equals $$\pi-\dfrac54$$, which corresponds to Option A.

Rewrite the linear combination of $$\sin x$$ and $$\cos x$$ in the form $$\cos(x-\alpha)$$.
Let $$\alpha$$ satisfy $$\cos\alpha=\frac{12}{13}$$ and $$\sin\alpha=\frac{5}{13}$$. Because $$\left(\frac{12}{13}\right)^2+\left(\frac{5}{13}\right)^2=\frac{144+25}{169}=1$$, such an $$\alpha$$ exists in the first quadrant.

Using the identity $$\cos(A-B)=\cos A\cos B+\sin A\sin B$$, put $$A=x,\;B=\alpha$$:
$$\cos(x-\alpha)=\cos x\cos\alpha+\sin x\sin\alpha$$ $$\Rightarrow\;\cos(x-\alpha)=\frac{12}{13}\cos x+\frac{5}{13}\sin x$$ $$-(1)$$

Hence $$\frac{12}{13}\cos x+\frac{5}{13}\sin x=\cos(x-\alpha)$$ with $$\alpha=\tan^{-1}\frac{5}{12}$$ $$-(2)$$

Now evaluate the required inverse cosine: $$\cos^{-1}\!\left(\frac{12}{13}\cos x+\frac{5}{13}\sin x\right)=\cos^{-1}\!\left(\cos(x-\alpha)\right)$$ $$-(3)$$

The principal value range of $$\cos^{-1}$$ is $$[\,0,\pi\,]$$. Given $$\dfrac{\pi}{2}\le x\le \dfrac{3\pi}{4}$$ and $$\alpha=\tan^{-1}\dfrac{5}{12}\approx0.394\text{ rad}$$, we obtain $$\frac{\pi}{2}-\alpha\le x-\alpha\le\frac{3\pi}{4}-\alpha$$ $$\Rightarrow\;1.18\text{ rad}\le x-\alpha\le1.96\text{ rad}$$

The interval $$[1.18,1.96]$$ lies completely inside $$[0,\pi]$$, so within this range $$\cos^{-1}\!\left(\cos(y)\right)=y$$ for $$y=x-\alpha$$.

Therefore $$\cos^{-1}\!\left(\frac{12}{13}\cos x+\frac{5}{13}\sin x\right)=x-\alpha$$ $$=x-\tan^{-1}\frac{5}{12}$$.

Hence the correct option is
Option C: $$x-\tan^{-1}\dfrac{5}{12}$$.

Using the identity $$\cot^{-1}(x) = \tan^{-1}(1/x)$$, the first term is:

$$\cot^{-1} \left( \frac{\beta(\alpha-\beta) + 1 + \beta^2}{\alpha-\beta} \right) = \cot^{-1} \left( \frac{\alpha\beta - \beta^2 + 1 + \beta^2}{\alpha-\beta} \right) = \tan^{-1} \left( \frac{\alpha-\beta}{1+\alpha\beta} \right)$$

This simplifies to: $$\tan^{-1}\alpha - \tan^{-1}\beta$$.

Term 2: $$\tan^{-1}\beta - \tan^{-1}\gamma$$

Term 3: $$\cot^{-1} \left( \frac{\alpha\gamma - \alpha^2 + 1 + \alpha^2}{\gamma - \alpha} \right) = \cot^{-1} \left( \frac{1+\alpha\gamma}{\gamma-\alpha} \right)$$

Note: Since $$\gamma - \alpha < 0$$, $$\cot^{-1}(x)$$ where $$x$$ is negative is $$\pi + \tan^{-1}(1/x)$$.

Term 3 $$= \pi + \tan^{-1} \left( \frac{\gamma-\alpha}{1+\gamma\alpha} \right) = \pi + \tan^{-1}\gamma - \tan^{-1}\alpha$$.

Sum: $$(\tan^{-1}\alpha - \tan^{-1}\beta) + (\tan^{-1}\beta - \tan^{-1}\gamma) + (\pi + \tan^{-1}\gamma - \tan^{-1}\alpha) = \mathbf{\pi}$$ (Option A).

For $$\cos^{-1} x$$: we need $$-1 \le x \le 1$$.

For $$\sin^{-1} x$$: we need $$-1 \le x \le 1$$.

For $$\sin^{-1}(2x+1)$$: we need $$-1 \le 2x+1 \le 1$$, i.e., $$-1 \le x \le 0$$.

So the domain is $$x \in [-1, 0]$$.

Substituting $$\cos^{-1} x = \dfrac{\pi}{2} - \sin^{-1} x$$:

$$\frac{\pi}{2} - \sin^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x+1)$$

$$-\frac{\pi}{2} = 2\sin^{-1} x + \sin^{-1}(2x+1)$$

$$\sin^{-1}(2x+1) = -\frac{\pi}{2} - 2\sin^{-1} x$$

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

Let $$\theta = \sin^{-1} x$$, so $$\sin\theta = x$$.

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

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

$$2x^2 - 2x - 2 = 0$$

$$x^2 - x - 1 = 0$$

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

$$x = \dfrac{1 + \sqrt{5}}{2} \approx 1.618$$ — outside the domain.

$$x = \dfrac{1 - \sqrt{5}}{2} \approx -0.618$$ — inside $$[-1, 0]$$. Valid.

For $$x = \dfrac{1-\sqrt{5}}{2}$$: we check that $$\sin^{-1}(2x+1) = \sin^{-1}(2-\sqrt{5})$$, and $$2-\sqrt{5} $$, which is in $$[-1, 1]$$, so the inverse sine is defined.

The RHS value $$-\frac{\pi}{2} - 2\sin^{-1}x $$ and $$\sin^{-1}(-0.236) $$

$$S = \left\{\dfrac{1-\sqrt{5}}{2}\right\}$$, so:

$$\sum_{x \in S}(2x-1)^2 = \left(2 \cdot \frac{1-\sqrt{5}}{2} - 1\right)^2 = (1 - \sqrt{5} - 1)^2 = (-\sqrt{5})^2 = 5$$

The answer is $$\boxed{5}$$.

We are given $$\alpha \le \beta$$, $$\alpha + \beta = 8$$, and $$\sec^2(\tan^{-1}\alpha) + \csc^2(\cot^{-1}\beta) = 36$$. We need to find $$\alpha^2 + \beta^2$$.

Simplify the trigonometric expressions.

Recall that $$\sec^2(\tan^{-1}\alpha) = 1 + \tan^2(\tan^{-1}\alpha) = 1 + \alpha^2$$

And $$\csc^2(\cot^{-1}\beta) = 1 + \cot^2(\cot^{-1}\beta) = 1 + \beta^2$$

(Using the identities $$\sec^2\theta = 1 + \tan^2\theta$$ and $$\csc^2\theta = 1 + \cot^2\theta$$)

Set up the equation.

$$(1 + \alpha^2) + (1 + \beta^2) = 36$$

$$\alpha^2 + \beta^2 + 2 = 36$$

$$\alpha^2 + \beta^2 = 34$$

The answer is 34.

Let $$\theta = \cos^{-1}\frac{x}{2}$$. By definition, $$\cos\theta = \frac{x}{2}$$ and $$-2 \le x \le 2$$.

Using $$\sin^2\theta + \cos^2\theta = 1$$ gives
$$\sin\theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\left(\frac{x}{2}\right)^2} = \frac{\sqrt{4 - x^{2}}}{2}\,.$$

The given function is $$y = \cos\left(\frac{\pi}{3} + \theta\right)$$. Apply the cosine addition formula
$$\cos(A+B) = \cos A \cos B - \sin A \sin B$$ with $$A = \frac{\pi}{3}$$ and $$B = \theta$$:

$$y = \cos\frac{\pi}{3}\,\cos\theta \;-\; \sin\frac{\pi}{3}\,\sin\theta.$$

Substitute $$\cos\frac{\pi}{3} = \frac12,\; \sin\frac{\pi}{3} = \frac{\sqrt3}{2},\; \cos\theta = \frac{x}{2},\; \sin\theta = \frac{\sqrt{4-x^{2}}}{2}$$:

$$y = \frac12 \cdot \frac{x}{2} \;-\; \frac{\sqrt3}{2} \cdot \frac{\sqrt{4-x^{2}}}{2} = \frac{x}{4} \;-\; \frac{\sqrt3}{4}\sqrt{4-x^{2}} = \frac14\Bigl(x - \sqrt3\,\sqrt{4-x^{2}}\Bigr).$$

Compute $$x - y$$:

$$x - y = x - \frac14\Bigl(x - \sqrt3\,\sqrt{4-x^{2}}\Bigr) = \frac34\,x + \frac{\sqrt3}{4}\sqrt{4-x^{2}} = \frac14\Bigl(3x + \sqrt3\,\sqrt{4-x^{2}}\Bigr).$$

Square both expressions.

Case 1: $$(x - y)^2$$ $$\begin{aligned} (x - y)^2 &= \left[\frac14\Bigl(3x + \sqrt3\,\sqrt{4-x^{2}}\Bigr)\right]^2 \\ &= \frac1{16}\Bigl(3x + \sqrt3\,\sqrt{4-x^{2}}\Bigr)^2 \\ &= \frac1{16}\Bigl(9x^{2} + 6\sqrt3\,x\,\sqrt{4-x^{2}} + 3(4 - x^{2})\Bigr) \\ &= \frac1{16}\Bigl(6x^{2} + 6\sqrt3\,x\,\sqrt{4-x^{2}} + 12\Bigr) \\ &= \frac{3}{8}\Bigl(x^{2} + \sqrt3\,x\,\sqrt{4-x^{2}} + 2\Bigr). \end{aligned}$$

Case 2: $$y^{2}$$ $$\begin{aligned} y^{2} &= \left[\frac14\Bigl(x - \sqrt3\,\sqrt{4-x^{2}}\Bigr)\right]^2 \\ &= \frac1{16}\Bigl(x - \sqrt3\,\sqrt{4-x^{2}}\Bigr)^2 \\ &= \frac1{16}\Bigl(x^{2} - 2\sqrt3\,x\,\sqrt{4-x^{2}} + 3(4 - x^{2})\Bigr) \\ &= \frac1{16}\Bigl(-2x^{2} - 2\sqrt3\,x\,\sqrt{4-x^{2}} + 12\Bigr). \end{aligned}$$

Multiply by 3:

$$3y^{2} = \frac{1}{16}\Bigl(-6x^{2} - 6\sqrt3\,x\,\sqrt{4-x^{2}} + 36\Bigr).$$

Add the two results:

$$\begin{aligned} (x - y)^2 + 3y^{2} &= \frac{3}{8}\Bigl(x^{2} + \sqrt3\,x\,\sqrt{4-x^{2}} + 2\Bigr) + \frac{1}{16}\Bigl(-6x^{2} - 6\sqrt3\,x\,\sqrt{4-x^{2}} + 36\Bigr) \\[4pt] &= \frac{6}{16}\Bigl(x^{2} + \sqrt3\,x\,\sqrt{4-x^{2}} + 2\Bigr) + \frac{1}{16}\Bigl(-6x^{2} - 6\sqrt3\,x\,\sqrt{4-x^{2}} + 36\Bigr) \\[4pt] &= \frac{6}{16}\Bigl(x^{2} + \sqrt3\,x\,\sqrt{4-x^{2}} + 2 - x^{2} - \sqrt3\,x\,\sqrt{4-x^{2}} + 6\Bigr) \\[4pt] &= \frac{6}{16}\Bigl(8\Bigr) = \frac{48}{16} = 3. \end{aligned}$$

Thus $$(x - y)^2 + 3y^2 = 3$$ for every $$x$$ in the domain $$[-2,2]$$.

Final Answer: 3

The series is

$$S=\cot^{-1}\!\left(\frac{7}{4}\right)+\cot^{-1}\!\left(\frac{19}{4}\right)+ \cot^{-1}\!\left(\frac{39}{4}\right)+\cot^{-1}\!\left(\frac{67}{4}\right)+\ldots$$

Step 1: Write the general term
The numerators form the sequence 7, 19, 39, 67, …
Differences: $$19-7=12,\;39-19=20,\;67-39=28,\ldots$$ These differences themselves form an A.P. with common difference 8. If $$k_n$$ is the  nth numerator, then

$$k_{n+1}=k_n+8n+4,\;k_1=7$$

Summing this recurrence gives

$$k_n=7+\sum_{m=1}^{\,n-1}(8m+4)=7+4(n-1)n+4(n-1)=4n^{2}+3$$

Hence the  nth term of the series is $$T_n=\cot^{-1}\!\left(\frac{4n^{2}+3}{4}\right)\qquad n=1,2,3,\ldots$$

Step 2: Convert $$\cot^{-1}$$ to $$\tan^{-1}$$
For positive arguments, $$\cot^{-1}x=\tan^{-1}\!\left(\frac1x\right)$$. Therefore

$$T_n=\tan^{-1}\!\left(\frac{4}{4n^{2}+3}\right)$$

Step 3: Express each term as a difference of two $$\tan^{-1}$$
Recall the identity $$\tan^{-1}A-\tan^{-1}B=\tan^{-1}\!\left(\frac{A-B}{1+AB}\right).$$ Choose $$A=\frac{2n+1}{2},\qquad B=\frac{2n-1}{2}.$$ Then $$A-B=\frac{2n+1-(2n-1)}{2}=1,$$ $$1+AB=1+\frac{(2n+1)(2n-1)}{4}=1+\frac{4n^{2}-1}{4}=\frac{4n^{2}+3}{4}.$$ Hence

$$\tan^{-1}A-\tan^{-1}B=\tan^{-1}\!\left(\frac{1}{(4n^{2}+3)/4}\right) =\tan^{-1}\!\left(\frac{4}{4n^{2}+3}\right)=T_n.$$

Thus

$$T_n=\tan^{-1}\!\left(\frac{2n+1}{2}\right)-\tan^{-1}\!\left(\frac{2n-1}{2}\right).$$

Step 4: Form the partial sum and telescope
Write the first  N  terms:

$$S_N=\sum_{n=1}^{N}T_n =\bigl[\tan^{-1}\!\left(\tfrac32\right)-\tan^{-1}\!\left(\tfrac12\right)\bigr] +\bigl[\tan^{-1}\!\left(\tfrac52\right)-\tan^{-1}\!\left(\tfrac32\right)\bigr] +\ldots +\bigl[\tan^{-1}\!\left(\tfrac{2N+1}{2}\right)-\tan^{-1}\!\left(\tfrac{2N-1}{2}\right)\bigr].$$

All intermediate terms cancel, leaving

$$S_N=\tan^{-1}\!\left(\frac{2N+1}{2}\right)-\tan^{-1}\!\left(\frac12\right).$$

Step 5: Take the limit as $$N\rightarrow\infty$$
$$\lim_{N\to\infty}\tan^{-1}\!\left(\frac{2N+1}{2}\right)=\tan^{-1}(\infty)=\frac{\pi}{2}.$$ Therefore

$$S=\lim_{N\to\infty}S_N=\frac{\pi}{2}-\tan^{-1}\!\left(\frac12\right).$$

Step 6: Match with the given options
Option D is $$\frac{\pi}{2}-\tan^{-1}\!\left(\frac12\right),$$ which equals the obtained sum.

Hence the required sum is $$\boxed{\frac{\pi}{2}-\tan^{-1}\!\left(\dfrac12\right)}$$, i.e. Option D.

Let $$t=\tan\theta$$. The given equation becomes

$$\theta=\tan^{-1}(2t)-\frac12\sin^{-1}\!\left(\frac{6t}{9+t^{2}}\right)\qquad -(1)$$

Step 1 : Simplify the $$\sin^{-1}$$ term.
Put $$t=3u$$ so that $$u=\dfrac{t}{3}$$. Then

$$\frac{6t}{9+t^{2}}=\frac{6\,(3u)}{9+9u^{2}} =\frac{18u}{9(1+u^{2})} =\frac{2u}{1+u^{2}} =\sin 2\alpha,$$
where $$\alpha=\tan^{-1}u=\tan^{-1}\!\left(\dfrac{t}{3}\right).$$

The principal value of $$\sin^{-1}(\sin 2\alpha)$$ equals $$2\alpha$$ whenever $$|2\alpha|\le\dfrac{\pi}{2}$$, i.e. $$|\alpha|\le\dfrac{\pi}{4}$$. Because $$|\alpha|\le\dfrac{\pi}{4}\iff |u|\le1\iff |t|\le3,$$ the replacement is valid for every $$t$$ that will survive the later algebra (it will turn out that $$|t|\le1$$).

Hence

$$\sin^{-1}\!\left(\frac{6t}{9+t^{2}}\right)=2\alpha =2\tan^{-1}\!\left(\frac{t}{3}\right).$$

Step 2 : Substitute back in equation $$(1)$$.

$$\theta=\tan^{-1}(2t)-\tan^{-1}\!\left(\frac{t}{3}\right)\qquad -(2)$$

Step 3 : Use the difference formula for $$\tan^{-1}$$.
For real $$a,b$$ within the principal interval,

$$\tan^{-1}a-\tan^{-1}b=\tan^{-1}\!\left(\frac{a-b}{1+ab}\right).$$

Applying this with $$a=2t,\; b=\dfrac{t}{3}$$ gives

$$\theta=\tan^{-1}\!\left(\frac{2t-\dfrac{t}{3}} {1+\dfrac{2t^{2}}{3}}\right) =\tan^{-1}\!\left(\frac{5t}{3+2t^{2}}\right)\qquad -(3)$$

Step 4 : Equate tangents.
Take tangent on both sides of $$(3)$$ (possible because both sides now lie in $$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$$):

$$t=\frac{5t}{3+2t^{2}}.$$

Multiply by $$3+2t^{2}$$:

$$t(3+2t^{2})=5t\quad\Longrightarrow\quad 3t+2t^{3}=5t\quad\Longrightarrow\quad 2t^{3}-2t=0.$$

Thus

$$t\bigl(t^{2}-1\bigr)=0\quad\Longrightarrow\quad t=0,\; t=1,\; t=-1.$$

These values satisfy $$|t|\le1\le3$$, so the earlier replacement in Step 1 is justified.

Step 5 : Find the corresponding $$\theta$$ values.

Step 6 : Count the solutions.
The equation is satisfied for exactly

$$\theta=-\frac{\pi}{4},\; 0,\; \frac{\pi}{4}.$$

Therefore the total number of real solutions is $$3$$.

Option C which is: 3

The argument of $$\sec^{-1}(u)$$ must satisfy $$|u| \ge 1$$.

So, $$|2[x]+1| \ge 1$$

Case 1: $$2[x]+1 \ge 1 \implies 2[x] \ge 0 \implies [x] \ge 0 \implies x \in [0, \infty)$$.

Case 2: $$2[x]+1 \le -1 \implies 2[x] \le -2 \implies [x] \le -1 \implies x \in (-\infty, 0)$$.

Combining these: $$(-\infty, 0) \cup [0, \infty) = (-\infty, \infty)$$

Let $$\alpha = \sin^{-1}\!\left(\frac{3}{5}\right)$$ and $$\beta = \cos^{-1}\!\left(\frac{2}{\sqrt{5}}\right)$$. Both are principal values, so $$-\frac{\pi}{2}\le \alpha\le \frac{\pi}{2}$$ and $$0\le \beta\le \pi$$.

Step 1: Evaluate $$\sin\alpha, \cos\alpha, \tan\alpha$$
Given $$\sin\alpha = \frac{3}{5}$$, hence $$\cos\alpha = \sqrt{1-\sin^2\alpha}= \sqrt{1-\left(\frac{3}{5}\right)^2}= \sqrt{\frac{16}{25}}=\frac{4}{5}$$ (positive in the principal range).
Therefore $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha}= \frac{\tfrac{3}{5}}{\tfrac{4}{5}} = \frac{3}{4}$$.

Step 2: Evaluate $$\sin\beta, \cos\beta, \tan\beta$$
Given $$\cos\beta = \frac{2}{\sqrt{5}}$$, so $$\sin\beta = \sqrt{1-\cos^2\beta}= \sqrt{1-\frac{4}{5}}= \sqrt{\frac{1}{5}}=\frac{1}{\sqrt{5}}$$ (positive because $$0\lt\beta\lt\frac{\pi}{2}$$).
Hence $$\tan\beta = \frac{\sin\beta}{\cos\beta}= \frac{\tfrac{1}{\sqrt{5}}}{\tfrac{2}{\sqrt{5}}}= \frac{1}{2}$$.

Step 3: Compute $$\tan(2\beta)$$
Use the double-angle formula $$\tan(2\beta)=\frac{2\tan\beta}{1-\tan^{2}\beta}$$.
Substituting $$\tan\beta=\frac{1}{2}$$ gives $$\tan(2\beta)=\frac{2\left(\frac{1}{2}\right)}{1-\left(\frac{1}{2}\right)^2}= \frac{1}{1-\frac{1}{4}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$$.

Step 4: Compute $$\tan\!\bigl(\alpha-2\beta\bigr)$$
The subtraction formula is $$\tan\!\bigl(\alpha-2\beta\bigr)=\frac{\tan\alpha-\tan(2\beta)}{1+\tan\alpha\,\tan(2\beta)}.$$

Numerator: $$\tan\alpha-\tan(2\beta)=\frac{3}{4}-\frac{4}{3}= \frac{9-16}{12}= -\frac{7}{12}.$$ Denominator: $$1+\tan\alpha\,\tan(2\beta)=1+\frac{3}{4}\cdot\frac{4}{3}=1+1=2.$$

Thus $$\tan\!\bigl(\alpha-2\beta\bigr)=\frac{-\tfrac{7}{12}}{2}= -\frac{7}{24}.$$

Therefore the required value is $$\frac{-7}{24}$$.

Option B which is: $$\frac{-7}{24}$$

Apply the formula $$\tan^{-1} A + \tan^{-1} B = \tan^{-1}(\frac{A+B}{1-AB})$$:

$$\tan^{-1}\left(\frac{x + 2x}{1 - x(2x)}\right) = \frac{\pi}{4} \implies \frac{3x}{1 - 2x^2} = \tan(\frac{\pi}{4}) = 1$$

$$3x = 1 - 2x^2 \implies 2x^2 + 3x - 1 = 0$$

Using the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$:

$$x = \frac{-3 \pm \sqrt{9 - 4(2)(-1)}}{4} = \frac{-3 \pm \sqrt{17}}{4}$$

Since the question asks for positive real values:

$$x = \frac{-3 + \sqrt{17}}{4}$$ (This is positive because $$\sqrt{17} > 3$$)

The other value is negative. Thus, there is only 1 positive solution

For $$f(x) = \sin^{-1}\left(\frac{x-1}{2x+3}\right)$$ to be defined, we need $$-1 \leq \frac{x-1}{2x+3} \leq 1$$ and $$2x+3 \neq 0$$.

Condition 1: $$\frac{x-1}{2x+3} \leq 1$$

$$\frac{x-1}{2x+3} - 1 \leq 0 \implies \frac{x-1-2x-3}{2x+3} \leq 0 \implies \frac{-x-4}{2x+3} \leq 0 \implies \frac{x+4}{2x+3} \geq 0$$

This holds when $$x \leq -4$$ or $$x > -\frac{3}{2}$$.

Condition 2: $$\frac{x-1}{2x+3} \geq -1$$

$$\frac{x-1}{2x+3} + 1 \geq 0 \implies \frac{x-1+2x+3}{2x+3} \geq 0 \implies \frac{3x+2}{2x+3} \geq 0$$

This holds when $$x \leq -\frac{3}{2}$$ or $$x \geq -\frac{2}{3}$$.

Intersection of both conditions:

$$\left((-\infty, -4] \cup (-\frac{3}{2}, \infty)\right) \cap \left((-\infty, -\frac{3}{2}) \cup [-\frac{2}{3}, \infty)\right)$$

$$= (-\infty, -4] \cup [-\frac{2}{3}, \infty)$$

So the domain is $$(-\infty, -4] \cup [-\frac{2}{3}, \infty) = \mathbb{R} - (-4, -\frac{2}{3})$$.

Thus $$\alpha = -4, \beta = -\frac{2}{3}$$.

$$12\alpha\beta = 12 \times (-4) \times (-\frac{2}{3}) = 32$$.

$$(\alpha + \beta + \gamma)(\alpha + \beta - \gamma) = 3\alpha\beta$$

$$(\alpha + \beta)^2 - \gamma^2 = 3\alpha\beta$$

$$\alpha^2 + 2\alpha\beta + \beta^2 - \gamma^2 = 3\alpha\beta$$

$$\alpha^2 - \alpha\beta + \beta^2 = \gamma^2 \quad \ldots (*)$$

The cosine rule states: $$c^2 = a^2 + b^2 - 2ab\cos C$$.

Comparing with (*) in the form $$\gamma^2 = \alpha^2 + \beta^2 - \alpha\beta$$, we get

$$2\alpha\beta \cos C = \alpha\beta \implies \cos C = \frac{1}{2} \implies C = \frac{\pi}{3}$$.

If $$\sin^{-1}\alpha + \sin^{-1}\beta + \sin^{-1}\gamma = \pi$$, let $$A = \sin^{-1}\alpha$$, $$B = \sin^{-1}\beta$$ and $$C = \sin^{-1}\gamma$$ so that $$A + B + C = \pi$$ and $$A, B, C$$ are the angles of a triangle. 

By the sine rule, the sides are proportional to the sines of the corresponding angles, namely $$\alpha, \beta, \gamma$$. Since the angle opposite the side of length $$\gamma$$ is $$C = \frac{\pi}{3}$$, it follows that $$\sin^{-1}\gamma = \frac{\pi}{3}$$.

$$\sin^{-1}\gamma = \frac{\pi}{3} \implies \gamma = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$$.

Let $$\cos^{-1} x = A \implies x = \cos A$$ and $$\sin^{-1} y = B \implies y = \sin B$$.

The given equation is $$A - B = \alpha \implies A = B + \alpha$$.

Substitute $$x$$ and $$y$$ into the expression:

$$E = \cos^2 A + \sin^2 B + 2 \cos A \sin B \sin \alpha$$

Replace $$A$$ with $$B + \alpha$$:

$$E = \cos^2(B + \alpha) + \sin^2 B + 2 \cos(B + \alpha) \sin B \sin \alpha$$

 Use the identity $$\cos(B + \alpha) = \cos B \cos \alpha - \sin B \sin \alpha$$:

$$E = (\cos B \cos \alpha - \sin B \sin \alpha)^2 + \sin^2 B + 2(\cos B \cos \alpha - \sin B \sin \alpha) \sin B \sin \alpha$$

Expanding and simplifying (specifically focusing on the $$\sin B \sin \alpha$$ terms cancelling out), the expression reduces to:

$$E = \cos^2 B \cos^2 \alpha + \sin^2 B \sin^2 \alpha - 2 \sin B \cos B \sin \alpha \cos \alpha + \sin^2 B + 2 \sin B \cos B \sin \alpha \cos \alpha - 2 \sin^2 B \sin^2 \alpha$$

$$E = \cos^2 B \cos^2 \alpha - \sin^2 B \sin^2 \alpha + \sin^2 B = \cos^2 B \cos^2 \alpha + \sin^2 B (1 - \sin^2 \alpha)$$

$$E = \cos^2 B \cos^2 \alpha + \sin^2 B \cos^2 \alpha = \cos^2 \alpha (\cos^2 B + \sin^2 B) = \cos^2 \alpha$$

The minimum value of $$\cos^2 \alpha$$ is 0 (which occurs when $$\alpha = \pi/2$$).

$$\sin^{-1}$$: range $$[-\pi/2, \pi/2]$$

$$\cos^{-1}$$: range $$[0, \pi]$$

 $$a = \sin^{-1}(\sin 5)$$

For $$x \in (3\pi/2, 2\pi)$$: $$\sin^{-1}(\sin x) = x - 2\pi$$.

$$a = 5 - 2\pi$$

 $$b = \cos^{-1}(\cos 5)$$. Since $$5\in$$ ($$3\pi/2,2\pi$$) :

For $$x \in (\pi, 2\pi)$$: $$\cos^{-1}(\cos x) = 2\pi - x$$.

$$b = 2\pi - 5$$

$$a^2 + b^2 = (5 - 2\pi)^2 + (2\pi - 5)^2 = 2(5 - 2\pi)^2$$

$$= 2(25 - 20\pi + 4\pi^2) = 50 - 40\pi + 8\pi^2$$

$$= 8\pi^2 - 40\pi + 50$$

1. Simplify the general term of the summation

Let us write the given expression as a summation from r = 1 to r = n:

$$L = \lim_{n \to \infty} \sum_{r=1}^{n} \left( \frac{n}{\sqrt{n^4+r^4}} - \frac{2n \cdot r^2}{(n^2+r^2)\sqrt{n^4+r^4}} \right)$$

We can combine the terms inside the summation by taking a common denominator:

$$\frac{n}{\sqrt{n^4+r^4}} \left( 1 - \frac{2r^2}{n^2+r^2} \right) = \frac{n}{\sqrt{n^4+r^4}} \left( \frac{n^2 + r^2 - 2r^2}{n^2+r^2} \right) = \frac{n(n^2 - r^2)}{(n^2+r^2)\sqrt{n^4+r^4}}$$

2. Express as a Riemann sum

To convert this into a definite integral, we factor out powers of n to express it in terms of \frac{r}{n}:

$$\frac{n \cdot n^2 (1 - \frac{r^2}{n^2})}{n^2 (1 + \frac{r^2}{n^2}) \cdot n^2 \sqrt{1 + \frac{r^4}{n^4}}} = \frac{1}{n} \cdot \frac{1 - \frac{r^2}{n^2}}{(1 + \frac{r^2}{n^2})\sqrt{1 + \frac{r^4}{n^4}}}$$

As $$n \to \infty, we replace \frac{1}{n}$$ with dx and $$\frac{r}{n}$$ with x. The limits of integration go from x = 0 to x = 1:

$$L = \int_0^1 \frac{1-x^2}{(1+x^2)\sqrt{1+x^4}} \, dx$$

3. Evaluate the definite integral

Divide both the numerator and the denominator of the integrand by $$x^2$$:

$$L = \int_0^1 \frac{\frac{1}{x^2}-1}{(x+\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}} \, dx$$

Let us use the substitution t = x + $$\frac{1}{x}$$. Differentiating both sides gives:

$$dt = \left(1 - \frac{1}{x^2}\right)dx \implies -dt = \left(\frac{1}{x^2} - 1\right)dx$$

We also express the term inside the square root in terms of t:

$$t^2 = x^2 + \frac{1}{x^2} + 2 \implies x^2 + \frac{1}{x^2} = t^2 - 2$$

Now find the new limits of integration:

When $$x \to 0^+, t \to \infty$$.

When x = 1, t = 1 + 1 = 2.

Substitute these into the integral:

$$L = \int_{\infty}^2 \frac{-dt}{t\sqrt{t^2-2}} = \int_2^{\infty} \frac{dt}{t\sqrt{t^2-2}}$$

4. Apply trigonometric substitution

Let t = $$\sqrt{2}\sec\theta$$, then dt = $$\sqrt{2}\sec\theta\tan\theta \, d\theta$$.

When t = 2, $$\sec\theta = \sqrt{2} \implies \theta = \frac{\pi}{4}$$.

When $$t \to \infty, \theta \to \frac{\pi}{2}$$.

Substitute these values into the integral:

$$L = \int_{\pi/4}^{\pi/2} \frac{\sqrt{2}\sec\theta\tan\theta \, d\theta}{\sqrt{2}\sec\theta \sqrt{2\sec^2\theta-2}} = \int_{\pi/4}^{\pi/2} \frac{\sqrt{2}\sec\theta\tan\theta \,d\theta}{\sqrt{2}\sec\theta \cdot \sqrt{2}\tan\theta}$$

Simplifying the integrand gives:

$$L = \int_{\pi/4}^{\pi/2} \frac{1}{\sqrt{2}} \, d\theta = \frac{1}{\sqrt{2}} \left[ \theta \right]_{\pi/4}^{\pi/2} = \frac{1}{\sqrt{2}} \left( \frac{\pi}{2} - \frac{\pi}{4} \right) =\frac{\pi}{4\sqrt{2}}$$

5. Determine the value of $$k^2$$

We are given that the limit value is equal to $$\frac{\pi}{k}$$:

$$\frac{\pi}{k} = \frac{\pi}{4\sqrt{2}} \implies k = 4\sqrt{2}$$

Squaring both sides gives:

$$k^2 = (4\sqrt{2})^2 = 16 \cdot 2 = 32$$

 $$\{x\} = x - \lfloor x \rfloor$$, so as $$x \to 0^+$$ we have $$\{x\} = x \to 0^+$$ while as $$x \to 0^-$$ we have $$\{x\} = x - (-1) = x + 1 \to 1^-$$.

To compute the right-hand limit $$R$$, set $$h = \{x\} = x \to 0^+$$ so that $$f = \frac{\cos^{-1}(1 - h^2)\sin^{-1}(1 - h)}{h - h^3} = \frac{\cos^{-1}(1 - h^2)\sin^{-1}(1 - h)}{h(1 - h^2)}$$. 

As $$h \to 0^+$$, we have $$\cos^{-1}(1 - h^2) \approx h\sqrt{2}$$ by using $$\cos^{-1}(1-u) \approx \sqrt{2u}$$ for small $$u = h^2$$, and $$\sin^{-1}(1 - h) \to \sin^{-1}(1) = \frac{\pi}{2}$$, while $$h(1 - h^2) \to h$$. 

Hence $$R = \lim_{h \to 0^+} \frac{h\sqrt{2}\cdot \frac{\pi}{2}}{h} = \frac{\pi\sqrt{2}}{2} = \frac{\pi}{\sqrt{2}}$$.

For the left-hand limit $$L$$, write $$h = \{x\} = x + 1 \to 1^-$$ in the form $$h = 1 - t$$ with $$t \to 0^+$$. 

Then $$1 - h^2 = t(2 - t)$$, $$1 - h = t$$, and $$h(1 - h^2) = (1-t)\,t\,(2-t)$$, so that $$f = \frac{\cos^{-1}(1 - h^2)\sin^{-1}(1 - h)}{h(1 - h^2)} = \frac{\cos^{-1}(t(2-t))\sin^{-1}(t)}{(1-t)\,t\,(2-t)}$$.

As $$t \to 0^+$$, we have $$\cos^{-1}(t(2-t)) \to \cos^{-1}(0) = \frac{\pi}{2}$$, $$\sin^{-1}(t) \approx t$$, and $$(1-t)\,t\,(2-t) \to 2t$$, giving $$L = \lim_{t \to 0^+} \frac{\frac{\pi}{2}\,t}{2t} = \frac{\pi}{4}$$.

Finally, since $$L^2 + R^2 = \frac{\pi^2}{16} + \frac{\pi^2}{2} = \frac{9\pi^2}{16}$$, we obtain $$\frac{32}{\pi^2}(L^2 + R^2) = \frac{32}{\pi^2} \cdot \frac{9\pi^2}{16} = 18$$.

We need to solve: $$2\sin^{-1}x + 3\cos^{-1}x = \frac{2\pi}{5}$$.

Using $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$, so $$\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$$.

Substituting: $$2\sin^{-1}x + 3\left(\frac{\pi}{2} - \sin^{-1}x\right) = \frac{2\pi}{5}$$

$$2\sin^{-1}x + \frac{3\pi}{2} - 3\sin^{-1}x = \frac{2\pi}{5}$$

$$-\sin^{-1}x = \frac{2\pi}{5} - \frac{3\pi}{2} = \frac{4\pi - 15\pi}{10} = -\frac{11\pi}{10}$$

$$\sin^{-1}x = \frac{11\pi}{10}$$

But the principal value of $$\sin^{-1}x$$ lies in $$[-\pi/2, \pi/2]$$, and $$\frac{11\pi}{10} > \frac{\pi}{2}$$. So there is no solution.

The number of real solutions is 0.

The answer is 0.

Convert to $$\tan^{-1}$$:

$$\tan^{-1}(\frac{1}{3}) + \tan^{-1}(\frac{1}{4}) + \tan^{-1}(\frac{1}{5}) + \tan^{-1}(\frac{1}{n}) = \frac{\pi}{4}$$

Combine first two terms:

$$\tan^{-1}(\frac{1/3 + 1/4}{1 - 1/12}) = \tan^{-1}(\frac{7/12}{11/12}) = \tan^{-1}(\frac{7}{11})$$

Combine with third term:

$$\tan^{-1}(\frac{7/11 + 1/5}{1 - 7/55}) = \tan^{-1}(\frac{46/55}{48/55}) = \tan^{-1}(\frac{23}{24})$$

Solve for $$n$$:

$$\tan^{-1}(\frac{23}{24}) + \tan^{-1}(\frac{1}{n}) = \tan^{-1}(1)$$

$$\tan^{-1}(\frac{1}{n}) = \tan^{-1}(1) - \tan^{-1}(\frac{23}{24}) = \tan^{-1}(\frac{1 - 23/24}{1 + 23/24}) = \tan^{-1}(\frac{1/24}{47/24}) = \tan^{-1}(\frac{1}{47})$$

Correct Answer: 47

Let us denote
$$\alpha=\tan^{-1}\!\left(\dfrac{6y}{\,9-y^{2}\,}\right),\qquad \beta=\cot^{-1}\!\left(\dfrac{9-y^{2}}{6y}\right)$$

According to the question, $$\alpha+\beta=\dfrac{2\pi}{3} \qquad\text{with}\qquad 0\lt |y|\lt 3$$

Because $$\beta$$ is a principal inverse-cotangent, $$\beta\in(0,\pi)$$, while $$\alpha\in\!\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$$. Put
$$u=\dfrac{6y}{\,9-y^{2}\,} \Longrightarrow \alpha=\tan^{-1}(u)$$ Then $$\dfrac{9-y^{2}}{6y}=\dfrac{1}{u}\,,$$ so $$\beta=\cot^{-1}\!\left(\dfrac{1}{u}\right).$$

Relationship between $$\alpha$$ and $$\beta$$
For every real $$u\neq0$$:

• If $$u\gt0$$, then $$\alpha\in(0,\pi/2)$$ and $$\beta=\tan^{-1}(1/u)\in(0,\pi/2).$$  Thus $$\beta=\alpha$$ and $$\alpha+\beta=2\alpha.$$
• If $$u\lt0$$, then $$\alpha\in(-\pi/2,0)$$.  Now $$\tan\beta=\dfrac{1}{u}\lt0$$, hence $$\beta$$ lies in the second quadrant, $$\beta=\pi-\tan^{-1}(|1/u|).$$  Because $$|\alpha|=\tan^{-1}(|u|)$$, we obtain $$\beta=\pi-|\alpha|=\pi+\,\alpha$$ and therefore $$\alpha+\beta=\pi+2\alpha.$$

Hence the given equation splits into two cases.

Case 1: $$u\gt0$$
$$2\alpha=\dfrac{2\pi}{3}\Longrightarrow \alpha=\dfrac{\pi}{3}$$ $$u=\tan\alpha=\sqrt{3}$$

Substituting $$u=\sqrt{3}$$:
$$\dfrac{6y}{\,9-y^{2}\,}=\sqrt{3}$$ $$6y=\sqrt{3}\,(9-y^{2})$$ $$\sqrt{3}\,y^{2}+6y-9\sqrt{3}=0$$ Dividing by $$\sqrt{3}$$ gives $$y^{2}+2\sqrt{3}\,y-9=0.$$

Solving: $$y=\dfrac{-2\sqrt{3}\pm4\sqrt{3}}{2}$$ $$\Rightarrow\;y=\sqrt{3}\quad\text{or}\quad y=-3\sqrt{3}.$$ Because $$0\lt|y|\lt3$$, only $$y=\sqrt{3}$$ is admissible.

Case 2: $$u\lt0$$
$$\pi+2\alpha=\dfrac{2\pi}{3}\Longrightarrow 2\alpha=-\dfrac{\pi}{3}\Longrightarrow \alpha=-\dfrac{\pi}{6}$$ $$u=\tan\alpha=-\dfrac{1}{\sqrt{3}}$$

Substituting $$u=-\dfrac{1}{\sqrt{3}}$$:
$$\dfrac{6y}{\,9-y^{2}\,}=-\dfrac{1}{\sqrt{3}}$$ Multiply by $$\sqrt{3}$$:
$$6\sqrt{3}\,y=-(9-y^{2})$$ $$y^{2}-6\sqrt{3}\,y-9=0$$

Discriminant $$\Delta=(6\sqrt{3})^{2}+36=108+36=144,\; \sqrt{\Delta}=12.$$ Thus $$y=\dfrac{6\sqrt{3}\pm12}{2}=3\sqrt{3}\pm6.$$

We get $$y_1=3\sqrt{3}+6\,(|y_1|>3,\;\text{rejected}),$$ $$y_2=3\sqrt{3}-6\,(\approx-0.804,\;|y_2|<3,\;\text{accepted}).$$

All acceptable solutions: $$y=\sqrt{3},\;y=3\sqrt{3}-6.$$

Required sum:
$$\sqrt{3}+\bigl(3\sqrt{3}-6\bigr)=4\sqrt{3}-6.$$

Therefore the desired value is $$4\sqrt{3}-6$$, i.e. Option C.

Given $$\sin^{-1}\frac{\alpha}{17} + \cos^{-1}\frac{4}{5} - \tan^{-1}\frac{77}{36} = 0$$ with $$0 < \alpha < 13$$, we need to find $$\sin^{-1}(\sin\alpha) + \cos^{-1}(\cos\alpha)$$.

To begin,

Rearranging: $$\sin^{-1}\frac{\alpha}{17} = \tan^{-1}\frac{77}{36} - \cos^{-1}\frac{4}{5}$$

Convert $$\cos^{-1}\frac{4}{5}$$ to inverse tangent. If $$\cos\theta = 4/5$$, then $$\sin\theta = 3/5$$ (for $$\theta \in [0, \pi/2]$$), so:

$$ \cos^{-1}\frac{4}{5} = \tan^{-1}\frac{3}{4} $$

Now use the subtraction formula $$\tan^{-1}a - \tan^{-1}b = \tan^{-1}\frac{a-b}{1+ab}$$ (valid when $$ab > -1$$):

$$ \tan^{-1}\frac{77}{36} - \tan^{-1}\frac{3}{4} = \tan^{-1}\left(\frac{\frac{77}{36} - \frac{3}{4}}{1 + \frac{77}{36} \cdot \frac{3}{4}}\right) $$

Numerator: $$\frac{77}{36} - \frac{3}{4} = \frac{77 - 27}{36} = \frac{50}{36} = \frac{25}{18}$$

Denominator: $$1 + \frac{77 \times 3}{36 \times 4} = 1 + \frac{231}{144} = \frac{144 + 231}{144} = \frac{375}{144} = \frac{125}{48}$$

$$ = \tan^{-1}\left(\frac{25/18}{125/48}\right) = \tan^{-1}\left(\frac{25}{18} \times \frac{48}{125}\right) = \tan^{-1}\left(\frac{1200}{2250}\right) = \tan^{-1}\frac{8}{15} $$

So $$\sin^{-1}\frac{\alpha}{17} = \tan^{-1}\frac{8}{15}$$. If $$\tan\theta = 8/15$$, then $$\sin\theta = \frac{8}{\sqrt{64+225}} = \frac{8}{\sqrt{289}} = \frac{8}{17}$$.

$$ \sin^{-1}\frac{\alpha}{17} = \sin^{-1}\frac{8}{17} \implies \alpha = 8 $$

Next,

Since $$\alpha = 8$$ radians, we need to use the principal value properties.

For $$\sin^{-1}(\sin 8)$$: Since $$8 \approx 2\pi + 1.717$$ and $$8 - 2\pi \approx 1.717$$, which lies in $$(\pi/2, \pi)$$. So $$\sin^{-1}(\sin 8) = \sin^{-1}(\sin(8 - 2\pi)) = \pi - (8 - 2\pi) = 3\pi - 8$$.

For $$\cos^{-1}(\cos 8)$$: Since $$8 \approx 2\pi + 1.717$$, and $$\cos(8) = \cos(8 - 2\pi)$$. Since $$8 - 2\pi \approx 1.717 \in (0, \pi)$$, we have $$\cos^{-1}(\cos 8) = 8 - 2\pi$$.

From this,

$$ \sin^{-1}(\sin 8) + \cos^{-1}(\cos 8) = (3\pi - 8) + (8 - 2\pi) = \pi $$

The answer is $$\pi$$.

The correct answer is Option 1: $$\pi$$.

Let $$x \in (0,1)$$ and set $$\alpha = \tan^{-1}x$$. Because $$0 \lt x \lt 1$$, we get $$0 \lt \alpha \lt \frac{\pi}{4}$$.

The given equation is
$$2 \tan^{-1}\!\left(\frac{1-x}{1+x}\right)=\cos^{-1}\!\left(\frac{1-x^{2}}{1+x^{2}}\right)\qquad -(1)$$

Step 1 : Simplify the left-hand side
Recall the identity $$\tan^{-1}\!\left(\frac{1-t}{1+t}\right)=\frac{\pi}{4}-\tan^{-1}t$$ for all $$t \gt 0$$.
Put $$t=x$$: then
$$\tan^{-1}\!\left(\frac{1-x}{1+x}\right)=\frac{\pi}{4}-\tan^{-1}x=\frac{\pi}{4}-\alpha.$$ Hence
$$\text{LHS}=2\Bigl(\frac{\pi}{4}-\alpha\Bigr)=\frac{\pi}{2}-2\alpha\qquad -(2)$$

Step 2 : Simplify the right-hand side
For $$\alpha=\tan^{-1}x$$ we have the double-angle identity
$$\cos 2\alpha=\frac{1-\tan^{2}\alpha}{1+\tan^{2}\alpha}=\frac{1-x^{2}}{1+x^{2}}.$$ Therefore
$$\text{RHS}=\cos^{-1}\!\left(\frac{1-x^{2}}{1+x^{2}}\right)=\cos^{-1}(\cos 2\alpha)\qquad -(3)$$

Step 3 : Evaluate $$\cos^{-1}(\cos 2\alpha)$$
The principal value of $$\cos^{-1}y$$ lies in $$[0,\pi]$$. Since $$0\lt\alpha\lt\frac{\pi}{4}$$, it follows that $$0\lt2\alpha\lt\frac{\pi}{2}$$ which is already inside $$[0,\pi]$$. Hence
$$\cos^{-1}(\cos 2\alpha)=2\alpha\qquad -(4)$$

Step 4 : Equate the two sides
From $$(2)$$ and $$(4)$$ substitute into $$(1)$$:
$$\frac{\pi}{2}-2\alpha = 2\alpha.$$ Solve for $$\alpha$$:
$$\frac{\pi}{2}=4\alpha \;\Longrightarrow\; \alpha=\frac{\pi}{8}.$$

Step 5 : Convert back to $$x$$
$$\alpha=\tan^{-1}x=\frac{\pi}{8}\;\Longrightarrow\; x=\tan\frac{\pi}{8}.$$ Using the known value $$\tan\frac{\pi}{8}=\sqrt{2}-1\approx0.414$$,
$$x=\sqrt{2}-1.$$

Step 6 : Verify range and count solutions
The obtained $$x$$ satisfies $$0\lt x\lt1$$, and $$x\approx0.414\lt\frac{1}{2}.$br We found exactly one solution.

Therefore $$n(S)=1$$ and that single element is less than $$$$\frac{1}{2}$$$$.
Hence, Option C is correct.

We need to evaluate $$\tan^{-1}\frac{1+\sqrt{3}}{3+\sqrt{3}} + \sec^{-1}\sqrt{\frac{8+4\sqrt{3}}{6+3\sqrt{3}}}$$.

Observe that $$\frac{1+\sqrt{3}}{3+\sqrt{3}} = \frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{1}{\sqrt{3}},$$ so $$\tan^{-1}\frac{1}{\sqrt{3}} = \frac{\pi}{6}$$.

Similarly, $$\frac{8+4\sqrt{3}}{6+3\sqrt{3}} = \frac{4(2+\sqrt{3})}{3(2+\sqrt{3})} = \frac{4}{3},$$ giving $$\sec^{-1}\sqrt{\frac{4}{3}} = \sec^{-1}\frac{2}{\sqrt{3}}$$. Since $$\sec\theta = \frac{2}{\sqrt{3}}$$ implies $$\cos\theta = \frac{\sqrt{3}}{2},$$ it follows that $$\theta = \frac{\pi}{6}$$.

Adding these two values yields $$\frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$$.

The answer is Option 3: $$\frac{\pi}{3}$$.

We need to find all solutions of $$\cos^{-1}(2x) - 2\cos^{-1}\sqrt{1-x^2} = \pi$$ for $$x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$$, and then evaluate $$\sum_{x \in S} 2\sin^{-1}(x^2 - 1)$$.

Determine the ranges of each term.

For $$x \in [-1/2, 1/2]$$:

- $$2x \in [-1, 1]$$, so $$\cos^{-1}(2x) \in [0, \pi]$$

- $$1 - x^2 \in [3/4, 1]$$, so $$\sqrt{1-x^2} \in [\sqrt{3}/2, 1]$$

- $$\cos^{-1}\sqrt{1-x^2} \in [0, \pi/6]$$

- Therefore $$2\cos^{-1}\sqrt{1-x^2} \in [0, \pi/3]$$

Analyze whether the equation can be satisfied.

The left-hand side equals:

$$ \text{LHS} = \cos^{-1}(2x) - 2\cos^{-1}\sqrt{1-x^2} $$

The maximum possible value of the LHS is achieved when $$\cos^{-1}(2x)$$ is maximum (= $$\pi$$, when $$x = -1/2$$) and $$2\cos^{-1}\sqrt{1-x^2}$$ is minimum (= 0, when $$x = 0$$). But these cannot occur simultaneously since $$x$$ cannot be both $$-1/2$$ and $$0$$.

Check specific values.

At $$x = -1/2$$:

$$ \cos^{-1}(2 \times (-1/2)) = \cos^{-1}(-1) = \pi $$

$$ \sqrt{1 - 1/4} = \sqrt{3}/2, \quad \cos^{-1}(\sqrt{3}/2) = \pi/6 $$

$$ \text{LHS} = \pi - 2 \times \pi/6 = \pi - \pi/3 = 2\pi/3 \neq \pi $$

At $$x = 0$$:

$$ \cos^{-1}(0) = \pi/2, \quad \sqrt{1-0} = 1, \quad \cos^{-1}(1) = 0 $$

$$ \text{LHS} = \pi/2 - 0 = \pi/2 \neq \pi $$

Show the LHS is always less than $$\pi$$.

Since $$\cos^{-1}(2x) \leq \pi$$ and $$2\cos^{-1}\sqrt{1-x^2} \geq 0$$, we have $$\text{LHS} \leq \pi$$. For equality, we need $$\cos^{-1}(2x) = \pi$$ (i.e., $$x = -1/2$$) AND $$\cos^{-1}\sqrt{1-x^2} = 0$$ (i.e., $$x = 0$$) simultaneously, which is impossible.

Moreover, at $$x = -1/2$$ (where the first term is maximized), the second term equals $$\pi/3 > 0$$, so $$\text{LHS} = \pi - \pi/3 = 2\pi/3 < \pi$$. Since the LHS is continuous and never reaches $$\pi$$, the equation has no solution in the given domain.

Evaluate the required sum.

Since the solution set $$S$$ is empty:

$$ \sum_{x \in S} 2\sin^{-1}(x^2 - 1) = 0 $$

The correct answer is Option 1: 0.

To solve this problem, we need to determine the value of $$\alpha$$ using the properties of the function $$g(x)$$ and then evaluate the inverse trigonometric expression.

Given:

$$g(x) = f(x) + f(1 - x)$$

To find the intervals of increase and decrease, we take the first derivative:

$$g'(x) = f'(x) + f'(1 - x) \cdot (-1) = f'(x) - f'(1 - x)$$

We are told $$g(x)$$ changes from decreasing to increasing at $$x = \alpha$$. This implies that $$x = \alpha$$ is a critical point where $$g'(\alpha) = 0$$:

$$f'(\alpha) - f'(1 - \alpha) = 0 \implies f'(\alpha) = f'(1 - \alpha)$$

Since $$f''(x) > 0$$ for $$x \in (0, 1)$$, the function $$f'(x)$$ is strictly increasing. In a strictly increasing function, if $$f'(a) = f'(b)$$, then $$a = b$$.

$$\alpha = 1 - \alpha \implies 2\alpha = 1 \implies \alpha = \frac{1}{2}$$

Now we substitute $$\alpha = \frac{1}{2}$$ into the required expression:

$$E = \tan^{-1}(2\alpha) + \tan^{-1}\left(\frac{1}{\alpha}\right) + \tan^{-1}\left(\frac{\alpha+1}{\alpha}\right)$$

Substitute $$\alpha = \frac{1}{2}$$:

• $$2\alpha = 2(1/2) = 1$$
• $$1/\alpha = 1/(1/2) = 2$$
• $$(\alpha+1)/\alpha = (1/2 + 1) / (1/2) = (3/2) / (1/2) = 3$$

The expression becomes:

$$E = \tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)$$

We know $$\tan^{-1}(1) = \frac{\pi}{4}$$. For the remaining terms, use the identity $$\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$$ (since $$xy > 1$$):

$$\tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1}\left(\frac{2+3}{1 - (2 \cdot 3)}\right)$$

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

$$\tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1}(-1)$$

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

Adding the first term back:

$$E = \frac{\pi}{4} + \frac{3\pi}{4} = \pi$$

Final Answer: A ($$\pi$$)

We need to find the domain of $$f(x) = \log_e\left(\frac{6x^2 + 5x + 1}{2x - 1}\right) + \cos^{-1}\left(\frac{2x^2 - 3x + 4}{3x - 5}\right)$$, expressed as $$(\alpha, \beta) \cup (\gamma, \delta)$$, and compute $$18(\alpha^2 + \beta^2 + \gamma^2 + \delta^2)$$.

Domain of the logarithm.

We need $$\frac{6x^2 + 5x + 1}{2x - 1} > 0$$.

Factoring the numerator: $$6x^2 + 5x + 1 = (2x + 1)(3x + 1)$$.

Critical points: $$x = -\frac{1}{2},\; -\frac{1}{3},\; \frac{1}{2}$$.

Sign analysis of $$\frac{(2x+1)(3x+1)}{2x-1}$$:

$$x < -\frac{1}{2}$$: $$\frac{(-)(-)}{(-)} = \frac{(+)}{(-)} < 0$$ ✘

$$-\frac{1}{2} < x < -\frac{1}{3}$$: $$\frac{(+)(-)}{(-)} = \frac{(-)}{(-)} > 0$$ ✔

$$-\frac{1}{3} < x < \frac{1}{2}$$: $$\frac{(+)(+)}{(-)} < 0$$ ✘

$$x > \frac{1}{2}$$: $$\frac{(+)(+)}{(+)} > 0$$ ✔

Domain from log: $$\left(-\frac{1}{2}, -\frac{1}{3}\right) \cup \left(\frac{1}{2}, \infty\right)$$

Domain of $$\cos^{-1}$$.

We need $$-1 \leq \frac{2x^2 - 3x + 4}{3x - 5} \leq 1$$ and $$x \neq \frac{5}{3}$$.

Let $$g(x) = \frac{2x^2 - 3x + 4}{3x - 5}$$.

Condition (i): $$g(x) \leq 1$$

$$\frac{2x^2 - 3x + 4 - (3x - 5)}{3x - 5} \leq 0 \implies \frac{2x^2 - 6x + 9}{3x - 5} \leq 0$$

The discriminant of $$2x^2 - 6x + 9$$ is $$36 - 72 = -36 < 0$$, and the leading coefficient is positive, so $$2x^2 - 6x + 9 > 0$$ for all $$x$$.

Therefore we need: $$3x - 5 < 0 \implies x < \frac{5}{3}$$.

Condition (ii): $$g(x) \geq -1$$

$$\frac{2x^2 - 3x + 4 + (3x - 5)}{3x - 5} \geq 0 \implies \frac{2x^2 - 1}{3x - 5} \geq 0$$

$$2x^2 - 1 = 0 \implies x = \pm\frac{1}{\sqrt{2}}$$.

Critical points: $$-\frac{1}{\sqrt{2}},\; \frac{1}{\sqrt{2}},\; \frac{5}{3}$$.

Sign analysis of $$\frac{2x^2 - 1}{3x - 5}$$:

$$x < -\frac{1}{\sqrt{2}}$$: $$\frac{(+)}{(-)} < 0$$ ✘

$$-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$$: $$\frac{(-)}{(-)} > 0$$ ✔

$$\frac{1}{\sqrt{2}} < x < \frac{5}{3}$$: $$\frac{(+)}{(-)} < 0$$ ✘

$$x > \frac{5}{3}$$: $$\frac{(+)}{(+)} > 0$$ ✔

Including zeros: $$\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] \cup \left(\frac{5}{3}, \infty\right)$$.

Combining conditions (i) and (ii): the $$\cos^{-1}$$ domain is $$\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$$.

Intersect with the log domain.

Log domain: $$\left(-\frac{1}{2}, -\frac{1}{3}\right) \cup \left(\frac{1}{2}, \infty\right)$$

$$\cos^{-1}$$ domain: $$\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$$

Since $$-\frac{1}{\sqrt{2}} \approx -0.707 < -\frac{1}{2}$$ and $$\frac{1}{\sqrt{2}} \approx 0.707$$:

$$\left(-\frac{1}{2}, -\frac{1}{3}\right) \cap \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] = \left(-\frac{1}{2}, -\frac{1}{3}\right)$$

$$\left(\frac{1}{2}, \infty\right) \cap \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] = \left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right)$$

At $$x = \frac{1}{\sqrt{2}}$$: $$g\left(\frac{1}{\sqrt{2}}\right) = \frac{1 - \frac{3}{\sqrt{2}} + 4}{\frac{3}{\sqrt{2}} - 5} = \frac{5 - \frac{3}{\sqrt{2}}}{\frac{3}{\sqrt{2}} - 5} = -1$$, and $$\cos^{-1}(-1) = \pi$$ is defined. However, the problem specifies open intervals $$(\alpha, \beta) \cup (\gamma, \delta)$$.

So the domain is: $$\left(-\frac{1}{2}, -\frac{1}{3}\right) \cup \left(\frac{1}{2}, \frac{1}{\sqrt{2}}\right)$$

Here $$\alpha = -\frac{1}{2},\; \beta = -\frac{1}{3},\; \gamma = \frac{1}{2},\; \delta = \frac{1}{\sqrt{2}}$$.

Compute the answer.

$$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = \frac{1}{4} + \frac{1}{9} + \frac{1}{4} + \frac{1}{2} = \frac{9 + 4 + 9 + 18}{36} = \frac{40}{36} = \frac{10}{9}$$

$$18\left(\alpha^2 + \beta^2 + \gamma^2 + \delta^2\right) = 18 \times \frac{10}{9} = 20$$

The answer is 20.

The given equation is $$\sqrt{1+\cos(2x)}=\sqrt{2}\,\tan^{-1}(\tan x)$$ to be solved for $$x$$ in
$$\left(-\tfrac{3\pi}{2},-\tfrac{\pi}{2}\right)\cup\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\cup\left(\tfrac{\pi}{2},\tfrac{3\pi}{2}\right).$$

Step 1 : Simplify the left-hand side
Using $$\cos(2x)=2\cos^{2}x-1,$$ we get $$1+\cos(2x)=2\cos^{2}x\; \Longrightarrow\; \sqrt{1+\cos(2x)}=\sqrt{2}\,|\,\cos x|.$$ Hence the equation becomes $$\sqrt{2}\,|\,\cos x|=\sqrt{2}\,\tan^{-1}(\tan x)$$ or after cancelling $$\sqrt{2}$$ $$|\,\cos x|=\tan^{-1}(\tan x). \quad -(1)$$

Step 2 : Evaluate $$\tan^{-1}(\tan x)$$ on the three sub-intervals
Because $$\tan^{-1}(\cdot)$$ returns values only in $$(-\tfrac{\pi}{2},\tfrac{\pi}{2}),$$ we have

Case 1: $$x\in\bigl(-\tfrac{\pi}{2},\tfrac{\pi}{2}\bigr)$$
 $$\tan^{-1}(\tan x)=x.$$ Case 2: $$x\in\bigl(\tfrac{\pi}{2},\tfrac{3\pi}{2}\bigr)$$
 Subtract $$\pi$$ to bring the angle into the principal range:  $$\tan^{-1}(\tan x)=x-\pi.$$ Case 3: $$x\in\bigl(-\tfrac{3\pi}{2},-\tfrac{\pi}{2}\bigr)$$
 Add $$\pi$$ for the same reason:  $$\tan^{-1}(\tan x)=x+\pi.$$

Step 3 : Solve equation (1) in each case

Case 1: $$|\,\cos x|=\cos x\;( \cos x\gt 0)\; \Longrightarrow\; \cos x = x,\quad x\in\bigl(-\tfrac{\pi}{2},\tfrac{\pi}{2}\bigr).$$
Define $$f(x)=\cos x-x.$$ $$f(0)=1\gt0$$ and $$f\!\bigl(\tfrac{\pi}{2}\bigr)=0-\tfrac{\pi}{2}\lt0;$$ $f(x)$$ is continuous and strictly decreasing on this interval, so exactly one root exists. This root lies near $$x$$\approx$$0.739$$ (between 0 and 1).

Case 2: Here $$|\,$$\cos$$ x|=-$$\cos$$ x\;( $$\cos$$ x\lt 0).$$ Equation (1) becomes $$-$$\cos$$ x = x-$$\pi$$,\quad x$$\in$$\bigl(\tfrac{$$\pi$$}{2},\tfrac{3$$\pi$$}{2}\bigr).$$ Put $$y=x-$$\pi$$\;( $$\Rightarrow$$ y$$\in$$(-\tfrac{$$\pi$$}{2},\tfrac{$$\pi$$}{2})).$$ Then $$-$$\cos$$(\,y+$$\pi$$)=y.$$ Because $$$$\cos$$(y+$$\pi$$)=-$$\cos$$ y,$$ this simplifies to $$$$\cos$$ y = y,\quad y$$\in$$\bigl(-\tfrac{$$\pi$$}{2},\tfrac{$$\pi$$}{2}\bigr).$$ This is identical to Case 1 and again has the single root $$y$$\approx$$0.739.$$ Therefore $$x=y+$$\pi$$$$\approx$$0.739+$$\pi$$$$\approx$$3.88,$$ which indeed lies in $$\bigl(\tfrac{$$\pi$$}{2},\tfrac{3$$\pi$$}{2}\bigr).$$

Case 3: Again $$|\,$$\cos$$ x|=-$$\cos$$ x$$ (cosine is negative in this interval). Equation (1) becomes $$-$$\cos$$ x = x+$$\pi$$,\quad x$$\in$$\bigl(-\tfrac{3$$\pi$$}{2},-\tfrac{$$\pi$$}{2}\bigr).$$ Set $$y=x+$$\pi$$\;( $$\Rightarrow$$ y$$\in$$(-\tfrac{$$\pi$$}{2},\tfrac{$$\pi$$}{2})).$$ With $$x=y-$$\pi$$,$$ we have $$-$$\cos$$(y-$$\pi$$)=y.$$ Since $$$$\cos$$(y-$$\pi$$)=-$$\cos$$ y,$$ this again reduces to $$$$\cos$$ y = y,\quad y$$\in$$\bigl(-\tfrac{$$\pi$$}{2},\tfrac{$$\pi$$}{2}\bigr).$$ We get the familiar single root $$y$$\approx$$0.739,$$ giving $$x=y-$$\pi$$$$\approx$$0.739-$$\pi$$$$\approx$$-2.40,$$ which is indeed in $$\bigl(-\tfrac{3$$\pi$$}{2},-\tfrac{$$\pi$$}{2}\bigr).$$

Step 4 : Count the solutions
Each case provides exactly one solution, and the three solutions are distinct: $$x_1$$\approx$$0.739,\; x_2$$\approx$$3.88,\; x_3$$\approx$$-2.40.$$ Hence the total number of real solutions is $$3.$$

Final Answer: 3

Given $$\tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{3}$$, with $$-1 < x < 1$$, $$x \neq 0$$.

Simplify using standard identities.

For $$|x| < 1$$: $$\tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}(x)$$.

For the $$\cot^{-1}$$ term, we use $$\cot^{-1}(y) = \frac{\pi}{2} - \tan^{-1}(y)$$.

Case 1 — $$0 < x < 1$$.

Here $$\frac{1-x^2}{2x} > 0$$. Note that $$\tan^{-1}\left(\frac{1-x^2}{2x}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{2}$$.

So $$\tan^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{2} - 2\tan^{-1}(x)$$.

Therefore $$\cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{2} - \left(\frac{\pi}{2} - 2\tan^{-1}(x)\right) = 2\tan^{-1}(x)$$.

The equation becomes:

$$2\tan^{-1}(x) + 2\tan^{-1}(x) = \frac{\pi}{3}$$

$$4\tan^{-1}(x) = \frac{\pi}{3} \implies \tan^{-1}(x) = \frac{\pi}{12}$$

$$x_1 = \tan\frac{\pi}{12} = 2 - \sqrt{3} \approx 0.268$$

Case 2 — $$-1 < x < 0$$.

Here $$\frac{1-x^2}{2x} < 0$$, and we need to be careful with the range of $$\cot^{-1}$$.

Writing $$x = -t$$ with $$t \in (0,1)$$: $$\frac{1-x^2}{2x} = -\frac{1-t^2}{2t}$$.

$$\cot^{-1}\left(-\frac{1-t^2}{2t}\right) = \pi - \cot^{-1}\left(\frac{1-t^2}{2t}\right) = \pi - 2\tan^{-1}(t) = \pi + 2\tan^{-1}(x)$$

The equation becomes:

$$2\tan^{-1}(x) + \pi + 2\tan^{-1}(x) = \frac{\pi}{3}$$

$$4\tan^{-1}(x) = -\frac{2\pi}{3} \implies \tan^{-1}(x) = -\frac{\pi}{6}$$

$$x_2 = -\tan\frac{\pi}{6} = -\frac{1}{\sqrt{3}} \approx -0.577$$

Compute the sum of solutions.

$$x_1 + x_2 = (2 - \sqrt{3}) - \frac{1}{\sqrt{3}} = 2 - \sqrt{3} - \frac{\sqrt{3}}{3} = 2 - \frac{3\sqrt{3} + \sqrt{3}}{3} = 2 - \frac{4\sqrt{3}}{3} = 2 - \frac{4}{\sqrt{3}}$$

Given this equals $$\alpha - \frac{4}{\sqrt{3}}$$, we get $$\alpha = 2$$.

The answer is $$2$$.

To find the value of the expression $$3\alpha + 10\beta + \gamma + 21\delta$$, we first need to determine the domain of the function $$f(x) = \sec^{-1} \left( \frac{2x}{5x+3} \right)$$.

The domain of $$\sec^{-1}(u)$$ is defined for $$|u| \geq 1$$. This implies:

$$\left| \frac{2x}{5x+3} \right| \geq 1$$

This absolute value inequality splits into two cases:

Case A: $$\frac{2x}{5x+3} \geq 1$$

Case B: $$\frac{2x}{5x+3} \leq -1$$

Subtract 1 from both sides:

$$\frac{2x}{5x+3} - 1 \geq 0 \implies \frac{2x - (5x+3)}{5x+3} \geq 0 \implies \frac{-3x - 3}{5x+3} \geq 0$$

Multiply by $$-1$$ (and flip the inequality sign):

$$\frac{3x+3}{5x+3} \leq 0 \implies \frac{x+1}{x + 3/5} \leq 0$$

Using the wavy curve method, the solution is $$x \in [-1, -3/5)$$.

Add 1 to both sides:

$$\frac{2x}{5x+3} + 1 \leq 0 \implies \frac{2x + (5x+3)}{5x+3} \leq 0 \implies \frac{7x+3}{5x+3} \leq 0$$

The critical points are $$x = -3/7$$ and $$x = -3/5$$.

Using the wavy curve method, the solution is $$x \in (-3/5, -3/7]$$.

The total domain is the union of the two cases:

$$[-1, -3/5) \cup (-3/5, -3/7]$$

Comparing this with the given format $$[\alpha, \beta) \cup (\gamma, \delta]$$, we identify:

• $$\alpha = -1$$
• $$\beta = -3/5$$
• $$\gamma = -3/5$$
• $$\delta = -3/7$$

Now, substitute these values into $$3\alpha + 10\beta + \gamma + 21\delta$$:

$$3(-1) + 10\left(-\frac{3}{5}\right) + \left(-\frac{3}{5}\right) + 21\left(-\frac{3}{7}\right)$$

$$-3 - 6 - \frac{3}{5} - 9$$

Combine the integers:

$$-18 - \frac{3}{5} = -\frac{90}{5} - \frac{3}{5} = -\frac{93}{5}$$

Let us denote

$$\theta = \cos^{-1}\!\Bigl(\sqrt{\frac{2}{2+\pi^{2}}}\Bigr).$$

Since $$0 \lt \sqrt{\dfrac{2}{2+\pi^{2}}} \lt 1$$, the principal value $$\theta$$ lies in $$\bigl(0,\tfrac{\pi}{2}\bigr).$$ Consequently

$$\cos\theta = \sqrt{\frac{2}{2+\pi^{2}}} \quad\Longrightarrow\quad \sin\theta = \sqrt{1-\cos^{2}\theta} = \sqrt{\frac{\pi^{2}}{2+\pi^{2}}} = \frac{\pi}{\sqrt{2+\pi^{2}}}.$$

1. First term

$$\frac{3}{2}\cos^{-1}\!\Bigl(\sqrt{\frac{2}{2+\pi^{2}}}\Bigr) = \frac{3}{2}\theta.$$

2. Second term Calculate the argument:

$$2\sqrt{2}\pi=\bigl(2\sin\theta\cos\theta\bigr)\!\bigl(2+\pi^{2}\bigr),$$ so

$$\frac{2\sqrt{2}\pi}{2+\pi^{2}} = 2\sin\theta\cos\theta = \sin(2\theta).$$

The principal range of $$\sin^{-1}x$$ is $$[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$$. Because $$\theta\in(0,\tfrac{\pi}{2})$$, we have $$2\theta\in\bigl(\tfrac{\pi}{2},\pi\bigr)$$, so $$\sin^{-1}\!\bigl(\sin(2\theta)\bigr)=\pi-2\theta.$$

Hence

$$\frac14\sin^{-1}\!\Bigl(\frac{2\sqrt{2}\pi}{2+\pi^{2}}\Bigr) =\frac14\bigl(\pi-2\theta\bigr) =\frac{\pi}{4}-\frac{\theta}{2}.$$

3. Third term Using $$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\pi}{\sqrt{2}},$$ we observe

$$\frac{\sqrt{2}}{\pi}=\cot\theta=\tan\!\bigl(\tfrac{\pi}{2}-\theta\bigr).$$

Since $$\tan^{-1}x$$ returns principal values in $$(-\tfrac{\pi}{2},\tfrac{\pi}{2})$$, we obtain

$$\tan^{-1}\!\Bigl(\frac{\sqrt{2}}{\pi}\Bigr)=\frac{\pi}{2}-\theta.$$

4. Adding the three terms

$$\begin{aligned} \text{Required value}&=\frac{3}{2}\theta +\Bigl(\frac{\pi}{4}-\frac{\theta}{2}\Bigr) +\Bigl(\frac{\pi}{2}-\theta\Bigr)\\[4pt] &=\Bigl(\frac{3}{2}\theta-\frac{\theta}{2}-\theta\Bigr) +\Bigl(\frac{\pi}{4}+\frac{\pi}{2}\Bigr)\\[4pt] &=0+\frac{3\pi}{4}=\frac{3\pi}{4}. \end{aligned}$$

Numerically, $$\dfrac{3\pi}{4}\approx2.35619$$ rad, which may be rounded as $$2.35$$ or $$2.36$$.

Answer : 2.35 | 2.36 (either value is accepted).

Let $$A=\alpha+\beta$$ and $$B=\alpha-\beta$$.
Given ranges  $$-\dfrac{\pi}{4}\lt\beta\lt 0\lt\alpha\lt\dfrac{\pi}{4}$$ imply $$0\lt A\lt\dfrac{\pi}{4}$$ and $$0\lt B\lt\dfrac{\pi}{2}$$, so the signs required below are fixed.

Data provided
$$\sin A=\dfrac13,\qquad\cos B=\dfrac23$$

1. Values for the individual functions of $$A$$ and $$B$$

$$\cos A=\sqrt{1-\sin ^2A}= \sqrt{1-\dfrac19}= \dfrac{2\sqrt2}{3}$$
$$\sin B=\sqrt{1-\cos ^2B}= \sqrt{1-\dfrac49}= \dfrac{\sqrt5}{3}$$

2. Values for $$\cos(A+B)$$ and $$\cos(A-B)$$ (addition-subtraction formulae)

$$\cos(A+B)=\cos A\cos B-\sin A\sin B =\dfrac{2\sqrt2}{3}\cdot\dfrac23-\dfrac13\cdot\dfrac{\sqrt5}{3} =\dfrac{4\sqrt2-\sqrt5}{9}\;.$$

$$\cos(A-B)=\cos A\cos B+\sin A\sin B =\dfrac{2\sqrt2}{3}\cdot\dfrac23+\dfrac13\cdot\dfrac{\sqrt5}{3} =\dfrac{4\sqrt2+\sqrt5}{9}\;.$$

3. Half-angle use to obtain $$\alpha=\dfrac{A+B}{2}$$ and $$\beta=\dfrac{A-B}{2}$$

Because $$0\lt\alpha\lt\dfrac{\pi}{4}$$ both $$\sin\alpha$$ and $$\cos\alpha$$ are positive.
Because $$-\dfrac{\pi}{4}\lt\beta\lt0$$ we have $$\cos\beta>0$$ but $$\sin\beta<0$$.

$$\sin\alpha =\sqrt{\dfrac{1-\cos2\alpha}{2}} =\sqrt{\dfrac{1-\cos(A+B)}{2}} =\sqrt{\dfrac{1-\dfrac{4\sqrt2-\sqrt5}{9}}{2}} \approx 0.5568$$

$$\cos\alpha =\sqrt{\dfrac{1+\cos2\alpha}{2}} =\sqrt{\dfrac{1+\dfrac{4\sqrt2-\sqrt5}{9}}{2}} \approx 0.8301$$

$$\sin\beta =-\sqrt{\dfrac{1-\cos2\beta}{2}} =-\sqrt{\dfrac{1-\cos(A-B)}{2}} =-\sqrt{\dfrac{1-\dfrac{4\sqrt2+\sqrt5}{9}}{2}} \approx -0.2480$$

$$\cos\beta =\sqrt{\dfrac{1+\cos2\beta}{2}} =\sqrt{\dfrac{1+\dfrac{4\sqrt2+\sqrt5}{9}}{2}} \approx 0.9687$$

4. Evaluate the four required quotients

$$\dfrac{\sin\alpha}{\cos\beta}\approx\dfrac{0.5568}{0.9687}=0.5748$$
$$\dfrac{\cos\beta}{\sin\alpha}\approx\dfrac{0.9687}{0.5568}=1.7388$$
$$\dfrac{\cos\alpha}{\sin\beta}\approx\dfrac{0.8301}{-0.2480}=-3.3489$$
$$\dfrac{\sin\beta}{\cos\alpha}\approx\dfrac{-0.2480}{0.8301}=-0.2988$$

Adding them:
$$S=\dfrac{\sin\alpha}{\cos\beta}+\dfrac{\cos\beta}{\sin\alpha} +\dfrac{\cos\alpha}{\sin\beta}+\dfrac{\sin\beta}{\cos\alpha} \approx 0.5748+1.7388-3.3489-0.2988 \approx -1.3341$$

5. Square of the sum

$$S^2\approx (-1.3341)^2\approx 1.78$$

6. The required greatest integer

The value $$S^2$$ lies between $$1$$ and $$2$$, so
$$\lfloor S^2\rfloor=1$$

Final answer: 1

Given code for the number of elements
$$P=2,\;Q=3,\;R=4,\;S=5,\;T=6$$

Case I : $$\cos x+\sin x=1,\;x\in\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]$$

Write $$\cos x+\sin x=\sqrt2\sin\left(x+\frac{\pi}{4}\right)$$.
Hence $$\sqrt2\sin\left(x+\frac{\pi}{4}\right)=1$$ gives
$$\sin\left(x+\frac{\pi}{4}\right)=\frac1{\sqrt2}\,.$$

For $$\sin\theta=\frac1{\sqrt2}$$, $$\theta=\frac{\pi}{4}+2k\pi$$ or $$\theta=\frac{3\pi}{4}+2k\pi$$.
So

$$x+\frac{\pi}{4}=\frac{\pi}{4}+2k\pi\;\Longrightarrow\;x=2k\pi$$
$$x+\frac{\pi}{4}=\frac{3\pi}{4}+2k\pi\;\Longrightarrow\;x=\frac{\pi}{2}+2k\pi$$

Inside $$\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]\;(=-120^\circ\text{ to }120^\circ)$$ we have
$$x=0,\;\;x=\frac{\pi}{2}$$ only. Thus there are $$2$$ solutions → code $$P$$.

Case II : $$\sqrt3\tan3x=1,\;x\in\left[-\frac{5\pi}{18},\frac{5\pi}{18}\right]$$

The equation is $$\tan3x=\frac1{\sqrt3}=\tan\frac{\pi}{6}$$.
Hence $$3x=\frac{\pi}{6}+k\pi\;\;(k\in\mathbb Z)$$ giving $$x=\frac{\pi}{18}+\frac{k\pi}{3}$$.

Check which integers $$k$$ keep $$x$$ in $$[-\frac{5\pi}{18},\,\frac{5\pi}{18}]$$:

k = -1 : $$x=-\frac{5\pi}{18}$$ (left end, allowed)
k = 0 : $$x=\frac{\pi}{18}$$ (inside interval)
k = 1 : $$x=\frac{7\pi}{18}$$ (>$$\frac{5\pi}{18}$$, rejected)
k ≤ -2 or k ≥ 2 give still larger magnitudes, rejected.

Thus exactly $$2$$ solutions → code $$P$$.

Case III : $$2\cos(2x)=\sqrt3,\;x\in\left[-\frac{6\pi}{5},\frac{6\pi}{5}\right]$$

Rewriting, $$\cos(2x)=\frac{\sqrt3}{2}=\cos\frac{\pi}{6}$$.
Therefore $$2x=2n\pi\pm\frac{\pi}{6}\;\Longrightarrow\;x=n\pi\pm\frac{\pi}{12}\;(n\in\mathbb Z).$$

Scan through the interval $$-\,\frac{6\pi}{5}(-216^\circ)\;\text{to}\;\frac{6\pi}{5}(216^\circ).$$

n = -1 : $$x=-\pi\pm\frac{\pi}{12}\;=\;-\,\frac{13\pi}{12},\;-\frac{11\pi}{12}$$ (both inside)
n = 0 : $$x=\pm\frac{\pi}{12}$$ (inside)
n = +1 : $$x=\pi\pm\frac{\pi}{12}\;=\;\frac{11\pi}{12},\;\frac{13\pi}{12}$$ (inside)
n = ±2 already lie outside ±$$\frac{6\pi}{5}$$.

Total number of solutions = $$2+2+2 = 6$$ → code $$T$$.

Case IV : $$\sin x-\cos x=1,\;x\in\left[-\frac{7\pi}{4},\frac{7\pi}{4}\right]$$

Write $$\sin x-\cos x=\sqrt2\sin\left(x-\frac{\pi}{4}\right)$$.
Thus $$\sqrt2\sin\left(x-\frac{\pi}{4}\right)=1$$ gives
$$\sin\left(x-\frac{\pi}{4}\right)=\frac1{\sqrt2}$$.

Hence

$$x-\frac{\pi}{4}=\frac{\pi}{4}+2k\pi\;\Longrightarrow\;x=\frac{\pi}{2}+2k\pi$$
$$x-\frac{\pi}{4}=\frac{3\pi}{4}+2k\pi\;\Longrightarrow\;x=\pi+2k\pi$$

Check inside $$-\,\frac{7\pi}{4}(-315^\circ)\;\text{to}\;\frac{7\pi}{4}(315^\circ):$$

k = -1 : $$x=-\frac{3\pi}{2},\;x=-\pi$$ (both inside)
k = 0 : $$x=\frac{\pi}{2},\;x=\pi$$ (inside)
k = 1 or k ≤ -2 give magnitudes >$$\frac{7\pi}{4}$$, rejected.

Total solutions = $$4$$ → code $$R$$.

Summary of matches
(I) → P  (II) → P  (III) → T  (IV) → R

Option B states exactly this mapping, so

Correct option: Option B which is: (I) → (P); (II) → (P); (III) → (T); (IV) → (R)

We are given the operation $$x \times y = x^2 + y^3$$, and the condition $$(x \times 1) \times 1 = x \times (1 \times 1)$$. We compute $$x \times 1 = x^2 + 1^3 = x^2 + 1$$ so the left side is $$(x^2 + 1) \times 1 = (x^2 + 1)^2 + 1^3 = (x^2 + 1)^2 + 1 = x^4 + 2x^2 + 1 + 1 = x^4 + 2x^2 + 2$$ while the right side is $$1 \times 1 = 1^2 + 1^3 = 2$$ and thus $$x \times 2 = x^2 + 2^3 = x^2 + 8$$. Setting these equal gives $$x^4 + 2x^2 + 2 = x^2 + 8$$ which simplifies to $$x^4 + x^2 - 6 = 0$$. Letting $$t = x^2$$ yields $$t^2 + t - 6 = 0$$, so $$(t + 3)(t - 2) = 0$$ and hence $$t = 2$$ (rejecting $$t = -3$$ since $$t = x^2 \ge 0$$), which means $$x^2 = 2$$ and $$x^4 = 4$$. Finally, the expression becomes $$\frac{x^4 + x^2 - 2}{x^4 + x^2 + 2} = \frac{4 + 2 - 2}{4 + 2 + 2} = \frac{4}{8} = \frac{1}{2}$$ and therefore $$2\sin^{-1}\left(\frac{1}{2}\right) = 2 \times \frac{\pi}{6} = \frac{\pi}{3}$$.

Therefore, the answer is Option B: $$\frac{\pi}{3}$$.

We need to evaluate:

$$\sin^{-1}\left(\sin\frac{2\pi}{3}\right) + \cos^{-1}\left(\cos\frac{7\pi}{6}\right) + \tan^{-1}\left(\tan\frac{3\pi}{4}\right)$$

Term 1: $$\sin^{-1}\left(\sin\frac{2\pi}{3}\right)$$

The range of $$\sin^{-1}$$ is $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$.

Since $$\frac{2\pi}{3}$$ is not in this range, we use: $$\sin\frac{2\pi}{3} = \sin\left(\pi - \frac{\pi}{3}\right) = \sin\frac{\pi}{3}$$

$$\sin^{-1}\left(\sin\frac{2\pi}{3}\right) = \frac{\pi}{3}$$

Term 2: $$\cos^{-1}\left(\cos\frac{7\pi}{6}\right)$$

The range of $$\cos^{-1}$$ is $$[0, \pi]$$.

Since $$\frac{7\pi}{6} > \pi$$, we use: $$\cos\frac{7\pi}{6} = \cos\left(2\pi - \frac{7\pi}{6}\right) = \cos\frac{5\pi}{6}$$

Since $$\frac{5\pi}{6} \in [0, \pi]$$:

$$\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}$$

Term 3: $$\tan^{-1}\left(\tan\frac{3\pi}{4}\right)$$

The range of $$\tan^{-1}$$ is $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$.

$$\tan\frac{3\pi}{4} = -1$$

$$\tan^{-1}(-1) = -\frac{\pi}{4}$$

Final Sum:

$$\frac{\pi}{3} + \frac{5\pi}{6} + \left(-\frac{\pi}{4}\right)$$

Taking LCM of 3, 6, and 4, which is 12:

$$= \frac{4\pi}{12} + \frac{10\pi}{12} - \frac{3\pi}{12}$$

$$= \frac{4\pi + 10\pi - 3\pi}{12}$$

$$= \frac{11\pi}{12}$$

The correct answer is Option A: $$\dfrac{11\pi}{12}$$.

We need to evaluate $$\tan\left(2\tan^{-1}\frac{1}{5} + \sec^{-1}\frac{\sqrt{5}}{2} + 2\tan^{-1}\frac{1}{8}\right)$$.

Using the double angle formula $$2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2}$$ (valid for $$|x| < 1$$):

$$2\tan^{-1}\frac{1}{5} = \tan^{-1}\frac{2 \cdot \frac{1}{5}}{1-\frac{1}{25}} = \tan^{-1}\frac{\frac{2}{5}}{\frac{24}{25}} = \tan^{-1}\frac{2 \times 25}{5 \times 24} = \tan^{-1}\frac{50}{120} = \tan^{-1}\frac{5}{12}$$

Let $$\alpha = \sec^{-1}\frac{\sqrt{5}}{2}$$. Then $$\sec\alpha = \frac{\sqrt{5}}{2}$$, so $$\cos\alpha = \frac{2}{\sqrt{5}}$$.

Using $$\sin^2\alpha = 1 - \cos^2\alpha = 1 - \frac{4}{5} = \frac{1}{5}$$, so $$\sin\alpha = \frac{1}{\sqrt{5}}$$.

Therefore $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{1/\sqrt{5}}{2/\sqrt{5}} = \frac{1}{2}$$.

So $$\sec^{-1}\frac{\sqrt{5}}{2} = \tan^{-1}\frac{1}{2}$$.

$$2\tan^{-1}\frac{1}{8} = \tan^{-1}\frac{2 \cdot \frac{1}{8}}{1-\frac{1}{64}} = \tan^{-1}\frac{\frac{1}{4}}{\frac{63}{64}} = \tan^{-1}\frac{64}{4 \times 63} = \tan^{-1}\frac{16}{63}$$

Using $$\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab}$$ (when $$ab < 1$$):

Here $$ab = \frac{5}{12} \times \frac{1}{2} = \frac{5}{24} < 1$$, so the formula applies.

$$\tan^{-1}\frac{5}{12} + \tan^{-1}\frac{1}{2} = \tan^{-1}\frac{\frac{5}{12}+\frac{1}{2}}{1-\frac{5}{24}} = \tan^{-1}\frac{\frac{5+6}{12}}{\frac{24-5}{24}} = \tan^{-1}\frac{\frac{11}{12}}{\frac{19}{24}} = \tan^{-1}\frac{11 \times 24}{12 \times 19} = \tan^{-1}\frac{22}{19}$$

Check: $$\frac{22}{19} \times \frac{16}{63} = \frac{352}{1197} < 1$$, so the addition formula applies.

Numerator: $$\frac{22}{19}+\frac{16}{63} = \frac{22 \times 63 + 16 \times 19}{19 \times 63} = \frac{1386 + 304}{1197} = \frac{1690}{1197}$$

Denominator: $$1-\frac{352}{1197} = \frac{1197-352}{1197} = \frac{845}{1197}$$

$$\tan\left(\tan^{-1}\frac{22}{19} + \tan^{-1}\frac{16}{63}\right) = \frac{1690}{845} = 2$$

Therefore, the correct answer is Option B: 2.

We need to evaluate $$\cos^{-1}\left(\frac{3}{10}\cos\left(\tan^{-1}\frac{4}{3}\right) + \frac{2}{5}\sin\left(\tan^{-1}\frac{4}{3}\right)\right)$$. First, let $$\theta = \tan^{-1}(4/3)$$ so that $$\tan\theta = 4/3$$. In a right triangle with opposite = 4, adjacent = 3, and hypotenuse = 5, it follows that $$\cos\theta = \frac{3}{5}$$ and $$\sin\theta = \frac{4}{5}$$.

Next, substituting these values into the expression gives $$\frac{3}{10}\cdot\frac{3}{5} + \frac{2}{5}\cdot\frac{4}{5} = \frac{9}{50} + \frac{8}{25} = \frac{9}{50} + \frac{16}{50} = \frac{25}{50} = \frac{1}{2}$$.

Now, since $$\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$$, the answer is Option C: $$\dfrac{\pi}{3}$$.

We need to evaluate $$\displaystyle\lim_{n \to \infty} 6\tan\left\{\sum_{r=1}^{n} \tan^{-1}\left(\frac{1}{r^2 + 3r + 3}\right)\right\}$$.

Simplify the general term using a telescoping identity:

Note that $$r^2 + 3r + 3 = 1 + (r + 1)(r + 2)$$.

Using the identity $$\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\!\left(\frac{A - B}{1 + AB}\right)$$, we observe:

$$\tan^{-1}(r + 2) - \tan^{-1}(r + 1) = \tan^{-1}\!\left(\frac{(r+2) - (r+1)}{1 + (r+1)(r+2)}\right) = \tan^{-1}\!\left(\frac{1}{1 + (r+1)(r+2)}\right)$$

$$= \tan^{-1}\!\left(\frac{1}{r^2 + 3r + 3}\right)$$

Evaluate the telescoping sum:

$$\sum_{r=1}^{n} \tan^{-1}\!\left(\frac{1}{r^2 + 3r + 3}\right) = \sum_{r=1}^{n} \left[\tan^{-1}(r + 2) - \tan^{-1}(r + 1)\right]$$

$$= \tan^{-1}(n + 2) - \tan^{-1}(2)$$

Take the limit as $$n \to \infty$$:

$$\lim_{n \to \infty} \left[\tan^{-1}(n + 2) - \tan^{-1}(2)\right] = \frac{\pi}{2} - \tan^{-1}(2)$$

Compute the final expression:

$$6\tan\!\left(\frac{\pi}{2} - \tan^{-1}(2)\right) = 6\cot\!\left(\tan^{-1}(2)\right) = 6 \times \frac{1}{2} = 3$$

The correct answer is Option C: $$3$$.

We need to find the sum of all solutions of $$\cos^{-1}(x) - 2\sin^{-1}(x) = \cos^{-1}(2x)$$.

Step 1: Use the identity $$\cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x)$$.

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

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

Step 2: Take cosine of both sides.

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

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

Step 3: Expand using the triple angle formula.

Let $$\theta = \sin^{-1}(x)$$, so $$\sin\theta = x$$.

$$\sin(3\theta) = 3\sin\theta - 4\sin^3\theta = 3x - 4x^3$$

Setting equal to $$2x$$:

$$3x - 4x^3 = 2x$$

$$x - 4x^3 = 0$$

$$x(1 - 4x^2) = 0$$

$$x = 0$$ or $$x^2 = \frac{1}{4}$$, i.e., $$x = \pm\frac{1}{2}$$

Step 4: Check domain constraints.

For $$\cos^{-1}(2x)$$ to be defined: $$-1 \leq 2x \leq 1 \Rightarrow -\frac{1}{2} \leq x \leq \frac{1}{2}$$.

Also, $$\cos^{-1}(x)$$ requires $$-1 \leq x \leq 1$$, which is already satisfied.

Additionally, LHS = $$\frac{\pi}{2} - 3\sin^{-1}(x)$$ must be in $$[0, \pi]$$ (range of $$\cos^{-1}$$).

Step 5: Verify each solution.

$$x = 0$$: LHS = $$\cos^{-1}(0) - 2\sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$. RHS = $$\cos^{-1}(0) = \frac{\pi}{2}$$. Valid. ✓

$$x = \frac{1}{2}$$: LHS = $$\cos^{-1}\left(\frac{1}{2}\right) - 2\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} - 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} - \frac{\pi}{3} = 0$$. RHS = $$\cos^{-1}(1) = 0$$. Valid. ✓

$$x = -\frac{1}{2}$$: LHS = $$\cos^{-1}\left(-\frac{1}{2}\right) - 2\sin^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} - 2\left(-\frac{\pi}{6}\right) = \frac{2\pi}{3} + \frac{\pi}{3} = \pi$$. RHS = $$\cos^{-1}(-1) = \pi$$. Valid. ✓

Step 6: Find the sum.

$$0 + \frac{1}{2} + \left(-\frac{1}{2}\right) = 0$$

The correct answer is Option A: $$0$$

Given $$0 < x < \frac{1}{\sqrt{2}}$$ and $$\frac{\sin^{-1}x}{\alpha} = \frac{\cos^{-1}x}{\beta}$$, we need to find the value of $$\sin\frac{2\pi\alpha}{\alpha + \beta}$$.

Let $$\frac{\sin^{-1}x}{\alpha} = \frac{\cos^{-1}x}{\beta} = k$$ (say)

Then $$\alpha = \frac{\sin^{-1}x}{k}$$ and $$\beta = \frac{\cos^{-1}x}{k}$$.

$$\alpha + \beta = \frac{\sin^{-1}x + \cos^{-1}x}{k} = \frac{\pi/2}{k}$$

(since $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$)

$$\frac{\alpha}{\alpha + \beta} = \frac{\sin^{-1}x / k}{\pi/(2k)} = \frac{2\sin^{-1}x}{\pi}$$

$$\frac{2\pi\alpha}{\alpha + \beta} = 2\pi \cdot \frac{\alpha}{\alpha + \beta} = 2\pi \cdot \frac{2\sin^{-1}x}{\pi} = 4\sin^{-1}x$$

Let $$\theta = \sin^{-1}x$$, so $$\sin\theta = x$$ and $$\cos\theta = \sqrt{1-x^2}$$.

$$\sin(4\theta) = 2\sin(2\theta)\cos(2\theta)$$

First: $$\sin(2\theta) = 2\sin\theta\cos\theta = 2x\sqrt{1-x^2}$$

Next: $$\cos(2\theta) = 1 - 2\sin^2\theta = 1 - 2x^2$$

Therefore:

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

This matches Option B.

Therefore, the correct answer is Option B: $$4x\sqrt{1-x^2}(1-2x^2)$$.

We need to find the domain of $$f(x) = \sin^{-1}\left(\dfrac{x^2 - 3x + 2}{x^2 + 2x + 7}\right)$$. For $$\sin^{-1}(u)$$ to be defined, we need $$-1 \leq u \leq 1$$, so we require $$-1 \leq \dfrac{x^2 - 3x + 2}{x^2 + 2x + 7} \leq 1$$.

We first note that the denominator $$x^2 + 2x + 7 = (x+1)^2 + 6 > 0$$ for all real $$x$$, so we can multiply inequalities without flipping signs.

Condition 1: $$\dfrac{x^2 - 3x + 2}{x^2 + 2x + 7} \leq 1$$. This gives $$x^2 - 3x + 2 \leq x^2 + 2x + 7$$, so $$-5x \leq 5$$, meaning $$x \geq -1$$.

Condition 2: $$\dfrac{x^2 - 3x + 2}{x^2 + 2x + 7} \geq -1$$. This gives $$x^2 - 3x + 2 \geq -(x^2 + 2x + 7) = -x^2 - 2x - 7$$, so $$2x^2 - x + 9 \geq 0$$. The discriminant is $$1 - 72 = -71 < 0$$, and since the leading coefficient is positive, this quadratic is always positive. So this condition is satisfied for all real $$x$$.

Combining both conditions, the domain is $$x \geq -1$$, i.e., $$[-1, \infty)$$.

Hence, the correct answer is Option C: $$[-1, \infty)$$.

We need to find all values of $$k$$ such that $$(\tan^{-1}x)^3 + (\cot^{-1}x)^3 = k\pi^3$$ has a solution for some $$x \in \mathbb{R}$$.

Let $$t = \tan^{-1}x$$. Then $$\cot^{-1}x = \frac{\pi}{2} - t$$, where $$t \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$.

The equation becomes:

$$ t^3 + \left(\frac{\pi}{2} - t\right)^3 = k\pi^3 $$

Expand using the identity $$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$$ where $$a = t$$ and $$b = \frac{\pi}{2} - t$$:

$$a + b = \frac{\pi}{2}$$

$$ a^2 - ab + b^2 = (a+b)^2 - 3ab = \frac{\pi^2}{4} - 3t\left(\frac{\pi}{2} - t\right) $$

$$ = \frac{\pi^2}{4} - \frac{3\pi t}{2} + 3t^2 $$

So:

$$ k\pi^3 = \frac{\pi}{2}\left(\frac{\pi^2}{4} - \frac{3\pi t}{2} + 3t^2\right) $$

$$ k\pi^3 = \frac{\pi^3}{8} - \frac{3\pi^2 t}{4} + \frac{3\pi t^2}{2} $$

Dividing by $$\pi^3$$:

$$ k = \frac{1}{8} - \frac{3t}{4\pi} + \frac{3t^2}{2\pi^2} $$

Let $$f(t) = \frac{3t^2}{2\pi^2} - \frac{3t}{4\pi} + \frac{1}{8}$$. This is a quadratic in $$t$$ opening upward.

The minimum occurs at:

$$ t = \frac{\frac{3}{4\pi}}{2 \cdot \frac{3}{2\pi^2}} = \frac{\frac{3}{4\pi}}{\frac{3}{\pi^2}} = \frac{3}{4\pi} \cdot \frac{\pi^2}{3} = \frac{\pi}{4} $$

Minimum value of $$k$$:

$$ k_{min} = \frac{3}{2\pi^2}\cdot\frac{\pi^2}{16} - \frac{3}{4\pi}\cdot\frac{\pi}{4} + \frac{1}{8} = \frac{3}{32} - \frac{3}{16} + \frac{1}{8} = \frac{3}{32} - \frac{6}{32} + \frac{4}{32} = \frac{1}{32} $$

Find the range. Since $$t \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ and $$f(t)$$ is a parabola opening upward with minimum at $$t = \frac{\pi}{4}$$:

At $$t = \frac{\pi}{4}$$: $$k = \frac{1}{32}$$ (achieved when $$x = \tan\frac{\pi}{4} = 1$$).

As $$t \to -\frac{\pi}{2}$$:

$$ k \to \frac{3}{2\pi^2}\cdot\frac{\pi^2}{4} + \frac{3}{4\pi}\cdot\frac{\pi}{2} + \frac{1}{8} = \frac{3}{8} + \frac{3}{8} + \frac{1}{8} = \frac{7}{8} $$

Since $$t = -\frac{\pi}{2}$$ is not attained (it would require $$x \to -\infty$$), the upper bound $$\frac{7}{8}$$ is not achieved.

At $$t = \frac{\pi}{4}$$ (i.e., $$x = 1$$), the minimum $$\frac{1}{32}$$ is achieved.

Therefore, the set of all values of $$k$$ is $$\left[\frac{1}{32}, \frac{7}{8}\right)$$.

The answer is Option A: $$\left[\dfrac{1}{32}, \dfrac{7}{8}\right)$$.

We need to find $$\cot\left(\sum_{n=1}^{50} \tan^{-1}\left(\frac{1}{1+n+n^2}\right)\right)$$. Note that $$1 + n + n^2 = 1 + n(1 + n)$$, so $$\frac{1}{1 + n + n^2} = \frac{(n+1) - n}{1 + n(n+1)}$$.

Using the inverse tangent subtraction formula $$\tan^{-1}A - \tan^{-1}B = \tan^{-1}\frac{A - B}{1 + AB}$$, we get $$\tan^{-1}\frac{(n+1) - n}{1 + n(n+1)} = \tan^{-1}(n+1) - \tan^{-1}(n)$$. Hence the sum telescopes:

$$\sum_{n=1}^{50} \tan^{-1}\left(\frac{1}{1+n+n^2}\right) = \sum_{n=1}^{50} [\tan^{-1}(n+1) - \tan^{-1}(n)] = \tan^{-1}(51) - \tan^{-1}(1) = \tan^{-1}(51) - \frac{\pi}{4}$$.

Let $$\theta = \tan^{-1}(51) - \frac{\pi}{4}$$. Then $$\tan\theta = \tan\left(\tan^{-1}(51) - \frac{\pi}{4}\right) = \frac{51 - 1}{1 + 51 \cdot 1} = \frac{50}{52} = \frac{25}{26}$$, so $$\cot\theta = \frac{1}{\tan\theta} = \frac{26}{25}$$.

The correct answer is Option C: $$\frac{26}{25}$$.

First evaluate the trigonometric numbers inside the inverse tangent.

Periodicity of cosine: $$\cos(\theta + 2\pi)=\cos\theta$$.
Write $$\frac{15\pi}{4}=2\pi+\frac{7\pi}{4}$$, so

$$\cos\frac{15\pi}{4}=\cos\frac{7\pi}{4}=\cos\left(-\frac{\pi}{4}\right)=\cos\frac{\pi}{4}.$$

Since $$\cos\frac{\pi}{4}=\frac{\sqrt2}{2}$$, the numerator becomes

$$\cos\frac{15\pi}{4}-1=\frac{\sqrt2}{2}-1.$$

The denominator is straightforward:

$$\sin\frac{\pi}{4}=\frac{\sqrt2}{2}.$$

Form the required fraction:

$$\frac{\cos\frac{15\pi}{4}-1}{\sin\frac{\pi}{4}} =\frac{\frac{\sqrt2}{2}-1}{\frac{\sqrt2}{2}} =\left(\frac{\sqrt2}{2}\right)\!\Big/\!\left(\frac{\sqrt2}{2}\right) -\;1\!\Big/\!\left(\frac{\sqrt2}{2}\right) =1-\frac{2}{\sqrt2} =1-\sqrt2.$$

Thus we need $$\tan^{-1}(1-\sqrt2).$$ Recall the well-known exact value

$$\tan\frac{\pi}{8}=\sqrt2-1.$$

Therefore

$$1-\sqrt2=-(\sqrt2-1)=-\tan\frac{\pi}{8} =\tan\!\left(-\frac{\pi}{8}\right).$$

Since the principal value range of $$\tan^{-1}$$ is $$(-\frac{\pi}{2},\frac{\pi}{2})$$, we obtain

$$\tan^{-1}(1-\sqrt2)=-\frac{\pi}{8}.$$

Hence the correct option is Option B: $$-$$\frac{\pi}{8}$$.$

We need to find the domain of $$f(x) = \cos^{-1}\left(\frac{x^2 - 4x + 2}{x^2 + 3}\right)$$.

For $$\cos^{-1}(u)$$ to be defined, we need $$-1 \leq u \leq 1$$.

Let $$g(x) = \frac{x^2 - 4x + 2}{x^2 + 3}$$. Note that $$x^2 + 3 > 0$$ for all $$x$$, so the denominator is always positive.

Condition 1: $$g(x) \leq 1$$

$$\frac{x^2 - 4x + 2}{x^2 + 3} \leq 1$$

$$x^2 - 4x + 2 \leq x^2 + 3$$

$$-4x \leq 1$$

$$x \geq -\frac{1}{4}$$

Condition 2: $$g(x) \geq -1$$

$$\frac{x^2 - 4x + 2}{x^2 + 3} \geq -1$$

$$x^2 - 4x + 2 \geq -(x^2 + 3)$$

$$2x^2 - 4x + 5 \geq 0$$

The discriminant is $$16 - 40 = -24 < 0$$, and the coefficient of $$x^2$$ is positive. So this inequality holds for all $$x \in \mathbb{R}$$.

Combining both conditions, the domain is $$\left[-\frac{1}{4}, \infty\right)$$.

The answer is Option B.

We evaluate each inverse trigonometric term separately.

Term 1: $$\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right)$$. Since $$\dfrac{2\pi}{3}$$ is not in $$\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$$, we use $$\sin\dfrac{2\pi}{3} = \sin\left(\pi - \dfrac{\pi}{3}\right) = \sin\dfrac{\pi}{3}$$. So $$\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right) = \dfrac{\pi}{3}$$.

Term 2: $$\cos^{-1}\!\left(\cos\dfrac{7\pi}{6}\right)$$. Since $$\dfrac{7\pi}{6}$$ is not in $$[0, \pi]$$, we use $$\cos\dfrac{7\pi}{6} = \cos\left(\pi + \dfrac{\pi}{6}\right) = -\cos\dfrac{\pi}{6}$$. Therefore $$\cos^{-1}\!\left(-\cos\dfrac{\pi}{6}\right) = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$$.

Term 3: $$\tan^{-1}\!\left(\tan\dfrac{3\pi}{4}\right)$$. Since $$\dfrac{3\pi}{4}$$ is not in $$\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$$, we use $$\tan\dfrac{3\pi}{4} = \tan\left(\pi - \dfrac{\pi}{4}\right) = -\tan\dfrac{\pi}{4} = -1$$. So $$\tan^{-1}(-1) = -\dfrac{\pi}{4}$$.

Adding all three terms:

$$\dfrac{\pi}{3} + \dfrac{5\pi}{6} + \left(-\dfrac{\pi}{4}\right) = \dfrac{4\pi}{12} + \dfrac{10\pi}{12} - \dfrac{3\pi}{12} = \dfrac{11\pi}{12}$$

The answer is Option B: $$\dfrac{11\pi}{12}$$.

We have $$x = \sin(2\tan^{-1}\alpha)$$ and $$y = \sin\left(\dfrac{1}{2}\tan^{-1}\dfrac{4}{3}\right)$$.

Using the identity $$\sin(2\tan^{-1}\alpha) = \dfrac{2\alpha}{1+\alpha^2}$$, it follows that

$$x = \frac{2\alpha}{1+\alpha^2}$$

Let $$\theta = \tan^{-1}\frac{4}{3}$$, so $$\tan\theta = \frac{4}{3}$$ and $$\cos\theta = \frac{3}{5}$$.

$$y = \sin\frac{\theta}{2} = \sqrt{\frac{1-\cos\theta}{2}} = \sqrt{\frac{1-\frac{3}{5}}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$$

$$y^2 = \frac{1}{5}$$

From $$y^2 = 1 - x$$ and the expressions above, we get

$$\frac{1}{5} = 1 - \frac{2\alpha}{1+\alpha^2}$$

$$\frac{2\alpha}{1+\alpha^2} = \frac{4}{5}$$

$$10\alpha = 4(1+\alpha^2) \implies 4\alpha^2 - 10\alpha + 4 = 0 \implies 2\alpha^2 - 5\alpha + 2 = 0$$

$$\alpha = \frac{5 \pm 3}{4} \implies \alpha = 2 \text{ or } \alpha = \frac{1}{2}$$

The required sum is

$$\sum_{\alpha \in S} 16\alpha^3 = 16\left(2^3 + \left(\frac{1}{2}\right)^3\right) = 16\left(8 + \frac{1}{8}\right) = 16 \times \frac{65}{8} = 130$$

The answer is $$\boxed{130}$$.

1. Initial State The eraser $$P$$ is at a fixed distance $$PC = \sqrt{5}$$ from the pivot $$C$$. The initial angle between the pencil and $$CP$$ is:$$\beta = \tan^{-1}(2)$$From the reference triangle:$$\tan \beta = 2$$ $$\sin \beta = \frac{2}{\sqrt{5}}$$ $$\cos \beta = \frac{1}{\sqrt{5}}$$

2. Final StateAfter rotating the pencil by an acute angle $$\alpha$$, the perpendicular distance from $$P$$ to the pencil is $$1$$ inch. Let the new angle between $$CP$$ and the pencil be $$\gamma$$.In the right-angled triangle formed:$$\sin \gamma = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{1}{\sqrt{5}}$$From this, we determine:$$\cos \gamma = \sqrt{1 - \sin^2 \gamma} = \sqrt{1 - \frac{1}{5}} = \frac{2}{\sqrt{5}}$$

$$\tan \gamma = \frac{\sin \gamma}{\cos \gamma} = \frac{1/\sqrt{5}}{2/\sqrt{5}} = \frac{1}{2}$$

3. Rotation CalculationFrom the diagram, the rotation angle $$\alpha$$ is the difference between the initial and final angles:$$\alpha = \beta - \gamma$$Apply the tangent subtraction identity:$$\tan \alpha = \tan(\beta - \gamma) = \frac{\tan \beta - \tan \gamma}{1 + \tan \beta \tan \gamma}$$Substitute the values:$$\tan \alpha = \frac{2 - \frac{1}{2}}{1 + (2 \cdot \frac{1}{2})} = \frac{1.5}{2} = \frac{3}{4}$$Final Answer:$$\alpha = \tan^{-1}\left(\frac{3}{4}\right)$$

We are given $$0 < a, b < 1$$ and $$\tan^{-1}a + \tan^{-1}b = \dfrac{\pi}{4}$$.

Using the addition formula, $$\tan\left(\tan^{-1}a + \tan^{-1}b\right) = \dfrac{a + b}{1 - ab} = \tan\dfrac{\pi}{4} = 1$$. Therefore $$a + b = 1 - ab$$, which gives $$a + b + ab = 1$$.

The given series is $$(a + b) - \dfrac{a^2 + b^2}{2} + \dfrac{a^3 + b^3}{3} - \dfrac{a^4 + b^4}{4} + \cdots$$

This can be written as $$\displaystyle\sum_{k=1}^{\infty}(-1)^{k+1}\dfrac{a^k + b^k}{k} = \displaystyle\sum_{k=1}^{\infty}(-1)^{k+1}\dfrac{a^k}{k} + \displaystyle\sum_{k=1}^{\infty}(-1)^{k+1}\dfrac{b^k}{k}$$.

Each sum is the Maclaurin series for $$\ln(1 + x)$$ evaluated at $$x = a$$ and $$x = b$$ respectively (valid since $$0 < a, b < 1$$). So the series equals $$\ln(1 + a) + \ln(1 + b) = \ln\big((1 + a)(1 + b)\big)$$.

Expanding: $$(1 + a)(1 + b) = 1 + a + b + ab = 1 + 1 = 2$$ (using $$a + b + ab = 1$$).

Therefore the value of the series is $$\ln 2 = \log_e 2$$.

We need to find $$\tan\left(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\right)$$.

Let $$\theta = \sin^{-1}\frac{\sqrt{63}}{8}$$, so $$\sin\theta = \frac{\sqrt{63}}{8}$$ and $$\cos\theta = \sqrt{1 - \frac{63}{64}} = \frac{1}{8}$$.

Using the half-angle formula, $$\cos\frac{\theta}{2} = \sqrt{\frac{1 + \cos\theta}{2}} = \sqrt{\frac{1 + \frac{1}{8}}{2}} = \sqrt{\frac{9}{16}} = \frac{3}{4}$$.

Similarly, $$\sin\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{2}} = \sqrt{\frac{1 - \frac{1}{8}}{2}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$.

Now we apply the half-angle formula again to find $$\tan\frac{\theta}{4}$$. Using $$\tan\frac{\alpha}{2} = \frac{1 - \cos\alpha}{\sin\alpha}$$ with $$\alpha = \frac{\theta}{2}$$, we get $$\tan\frac{\theta}{4} = \frac{1 - \cos\frac{\theta}{2}}{\sin\frac{\theta}{2}} = \frac{1 - \frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{\frac{1}{4}}{\frac{\sqrt{7}}{4}} = \frac{1}{\sqrt{7}}$$.

Therefore, a possible value of $$\tan\left(\frac{1}{4}\sin^{-1}\frac{\sqrt{63}}{8}\right)$$ is $$\frac{1}{\sqrt{7}}$$.

We need to find the number of real solutions of $$\sin^{-1}\left(\frac{3x}{5}\right) + \sin^{-1}\left(\frac{4x}{5}\right) = \sin^{-1}(x)$$.

For the equation to be well-defined, we need $$\left|\frac{3x}{5}\right| \leq 1$$, $$\left|\frac{4x}{5}\right| \leq 1$$, and $$|x| \leq 1$$. The most restrictive condition is $$|x| \leq 1$$.

Clearly, $$x = 0$$ satisfies the equation since all three terms become 0.

For $$x \neq 0$$, we rearrange: $$\sin^{-1}\left(\frac{3x}{5}\right) = \sin^{-1}(x) - \sin^{-1}\left(\frac{4x}{5}\right)$$. Taking the sine of both sides and applying the identity $$\sin(A - B) = \sin A \cos B - \cos A \sin B$$:

$$\frac{3x}{5} = x\sqrt{1 - \frac{16x^2}{25}} - \frac{4x}{5}\sqrt{1 - x^2}$$

Since $$x \neq 0$$, dividing both sides by $$x$$ (noting the sign is preserved since sine inverse is odd):

$$\frac{3}{5} = \frac{1}{5}\sqrt{25 - 16x^2} - \frac{4}{5}\sqrt{1 - x^2}$$

Multiplying by 5: $$3 = \sqrt{25 - 16x^2} - 4\sqrt{1 - x^2}$$.

Let $$u = \sqrt{1 - x^2} \geq 0$$. Then $$25 - 16x^2 = 25 - 16(1 - u^2) = 9 + 16u^2$$. The equation becomes:

$$\sqrt{9 + 16u^2} = 3 + 4u$$

The right side must be non-negative, which holds since $$u \geq 0$$. Squaring both sides (valid since both sides are non-negative):

$$9 + 16u^2 = 9 + 24u + 16u^2$$

This simplifies to $$24u = 0$$, so $$u = 0$$, which gives $$1 - x^2 = 0$$, i.e., $$x = \pm 1$$.

Verification for $$x = 1$$: $$\sin^{-1}\left(\frac{3}{5}\right) + \sin^{-1}\left(\frac{4}{5}\right)$$. Since $$\sin^{-1}\left(\frac{3}{5}\right)$$ and $$\sin^{-1}\left(\frac{4}{5}\right)$$ are complementary angles (as $$3, 4, 5$$ form a Pythagorean triple, $$\cos\left(\sin^{-1}\frac{3}{5}\right) = \frac{4}{5}$$), their sum equals $$\frac{\pi}{2} = \sin^{-1}(1)$$. Valid.

Verification for $$x = -1$$: By the odd symmetry of $$\sin^{-1}$$, $$\sin^{-1}\left(-\frac{3}{5}\right) + \sin^{-1}\left(-\frac{4}{5}\right) = -\frac{\pi}{2} = \sin^{-1}(-1)$$. Valid.

Therefore, there are 3 real values of $$x$$: $$x \in \{-1, 0, 1\}$$.

We have to compute the expression $$$\cos^{-1}(\cos(-5))+\sin^{-1}(\sin 6)-\tan^{-1}(\tan 12)$$$ where every inverse trigonometric function is to be taken in its principal value branch.

First, recall the principal value ranges:

$$$\cos^{-1}x\in[0,\pi],\qquad \sin^{-1}x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right],\qquad \tan^{-1}x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right).$$$

1. Evaluating $$\cos^{-1}(\cos(-5))$$

Because the cosine function is even, $$\cos(-5)=\cos 5.$$ However, the principal value of $$\cos^{-1}$$ must lie in $$[0,\pi].$$ Now $$5$$ radians exceeds $$\pi$$ (since $$\pi\approx3.14$$) but is less than $$2\pi$$. Using the identity $$\cos\theta=\cos(2\pi-\theta),$$ we write

$$\cos 5=\cos(2\pi-5).$$

The angle $$2\pi-5$$ clearly lies between $$0$$ and $$\pi$$ because $$0<2\pi-5<\pi.$$ Therefore

$$\cos^{-1}(\cos(-5))=2\pi-5.$$

2. Evaluating $$\sin^{-1}(\sin 6)$$

The angle $$6$$ radians is slightly less than $$2\pi$$, so it is in the interval $$(\pi,2\pi).$$ For such angles the sine is negative and we use the fact that

$$\sin\theta=\sin(\theta-2\pi).$$

Subtracting one full revolution gives

$$6-2\pi\;(\approx -0.283)\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right].$$

Since this result already lies in the principal value range of $$\sin^{-1},$$ we have

$$\sin^{-1}(\sin 6)=6-2\pi.$$

3. Evaluating $$\tan^{-1}(\tan 12)$$

The tangent function has period $$\pi,$$ so we may subtract integer multiples of $$\pi$$ until we land in the interval $$\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right).$$ Write $$12=n\pi+\alpha$$ with $$\alpha\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right).$$

Dividing by $$\pi$$ we find $$\dfrac{12}{\pi}\approx3.82,$$ so choose $$n=4.$$ Then

$$\alpha=12-4\pi.$$

Numerically $$12-4\pi\approx -0.566,$$ which indeed lies between $$-\dfrac{\pi}{2}$$ and $$\dfrac{\pi}{2}.$$ Thus

$$$\tan 12=\tan(12-4\pi)\quad\text{and}\quad\tan^{-1}(\tan 12)=12-4\pi.$$$

4. Combining all three parts

Substituting the three principal values we obtained:

$$$\begin{aligned} \cos^{-1}(\cos(-5))+\sin^{-1}(\sin 6)-\tan^{-1}(\tan 12) &=(2\pi-5)+(6-2\pi)-(12-4\pi)\\ &=2\pi-5+6-2\pi-12+4\pi\\ &=(2\pi-2\pi+4\pi)+(-5+6-12)\\ &=4\pi-11. \end{aligned}$$$

Hence, the correct answer is Option C.

For the equation $$\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2+x+1} = \frac{\pi}{4}$$, we first determine the domain.

For $$\tan^{-1}\sqrt{x(x+1)}$$ to be defined (and real-valued with the argument non-negative), we need $$x(x+1) \ge 0$$, so $$x \le -1$$ or $$x \ge 0$$.

For $$\sin^{-1}\sqrt{x^2+x+1}$$ to be defined, we need $$0 \le x^2+x+1 \le 1$$, i.e., $$x^2+x \le 0$$, which means $$-1 \le x \le 0$$.

Combining both conditions, the domain is $$\{-1, 0\}$$.

Now note that $$x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \ge \frac{3}{4}$$ for all real $$x$$. On the domain $$\{-1, 0\}$$: at $$x = 0$$, $$x^2+x+1 = 1$$; at $$x = -1$$, $$x^2+x+1 = 1$$.

At $$x = 0$$: $$\tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2} \ne \frac{\pi}{4}$$.

At $$x = -1$$: $$\tan^{-1}(0) + \sin^{-1}(1) = 0 + \frac{\pi}{2} = \frac{\pi}{2} \ne \frac{\pi}{4}$$.

Neither point in the domain satisfies the equation. Furthermore, since $$\sin^{-1}\sqrt{x^2+x+1} \ge \sin^{-1}(1) = \frac{\pi}{2}$$ at the boundary points (where $$x^2+x=0$$, giving the value 1 under the square root), the left side is always at least $$\frac{\pi}{2} \gt \frac{\pi}{4}$$ throughout the domain.

Therefore the equation has $$0$$ real roots.

Let $$\alpha = 2\tan^{-1}\!\left(\frac{3}{5}\right)$$ and $$\beta = \sin^{-1}\!\left(\frac{5}{13}\right)$$. We compute $$\tan(\alpha + \beta)$$.

Using the double-angle formula: $$\tan\alpha = \tan\!\left(2\tan^{-1}\frac{3}{5}\right) = \frac{2 \cdot \frac{3}{5}}{1 - \left(\frac{3}{5}\right)^2} = \frac{\frac{6}{5}}{\frac{16}{25}} = \frac{6}{5} \cdot \frac{25}{16} = \frac{15}{8}.$$

For $$\beta = \sin^{-1}\!\left(\frac{5}{13}\right)$$: $$\sin\beta = \frac{5}{13}$$, so $$\cos\beta = \frac{12}{13}$$ and $$\tan\beta = \frac{5}{12}$$.

Applying the addition formula: $$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{\frac{15}{8} + \frac{5}{12}}{1 - \frac{15}{8} \cdot \frac{5}{12}}.$$

$$\text{Numerator: } \frac{15}{8} + \frac{5}{12} = \frac{45}{24} + \frac{10}{24} = \frac{55}{24}.$$ $$\text{Denominator: } 1 - \frac{75}{96} = \frac{21}{96} = \frac{7}{32}.$$

$$\tan(\alpha + \beta) = \frac{55}{24} \div \frac{7}{32} = \frac{55}{24} \cdot \frac{32}{7} = \frac{55 \cdot 4}{3 \cdot 7} = \frac{220}{21}.$$

We have to evaluate the expression

$$p=\sum_{r=1}^{50}\tan^{-1}\frac{1}{2r^{2}}$$

and then obtain $$\tan p.$$

First, recall the subtraction formula for the inverse tangent:

$$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\!\left(\frac{x-y}{1+xy}\right).$$

We want the right-hand side of this identity to be $$\tan^{-1}\dfrac{1}{2r^{2}}.$$ To achieve that, choose

$$x=2r+1,\qquad y=2r-1.$$

Substituting these in the fraction that appears in the formula gives

$$$\frac{x-y}{1+xy}= \frac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)} =\frac{2}{1+\left((2r)^{2}-1\right)} =\frac{2}{1+4r^{2}-1} =\frac{2}{4r^{2}} =\frac{1}{2r^{2}}.$$$

Hence

$$$\tan^{-1}\frac{1}{2r^{2}} =\tan^{-1}(2r+1)-\tan^{-1}(2r-1).$$$

We now replace every term in the given sum by this difference:

$$$\begin{aligned} p&=\sum_{r=1}^{50}\left[\tan^{-1}(2r+1)-\tan^{-1}(2r-1)\right]\\[4pt] &=\bigl(\tan^{-1}3-\tan^{-1}1\bigr) +\bigl(\tan^{-1}5-\tan^{-1}3\bigr) +\bigl(\tan^{-1}7-\tan^{-1}5\bigr)\\[2pt] &\ \ \,+\;\cdots\; +\bigl(\tan^{-1}101-\tan^{-1}99\bigr). \end{aligned}$$$

Observe the telescoping nature of this series: every intermediate term cancels with its negative counterpart. All that survives is the first negative term and the last positive term. Therefore

$$p=\tan^{-1}101-\tan^{-1}1.$$

To find $$\tan p$$ we again use the tangent-difference formula, this time for ordinary tangents:

$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\,\tan B}.$$

Here $$A=\tan^{-1}101$$ and $$B=\tan^{-1}1,$$ so $$\tan A=101$$ and $$\tan B=1.$$ Substituting, we get

$$$\tan p =\frac{101-1}{1+101\cdot1} =\frac{100}{102} =\frac{50}{51}.$$$

Hence, the correct answer is Option C.

We need to evaluate $$S_k = \sum_{r=1}^{k} \tan^{-1}\left(\frac{6^r}{2^{2r+1} + 3^{2r+1}}\right)$$ and find $$\lim_{k \to \infty} S_k$$.

We write $$6^r = 2^r \cdot 3^r$$ and note that $$2^{2r+1} + 3^{2r+1} = 2 \cdot 4^r + 3 \cdot 9^r$$. We claim each term telescopes as $$\tan^{-1}\left(\frac{3}{2}\right)^{r+1} - \tan^{-1}\left(\frac{3}{2}\right)^r$$.

To verify, we use the subtraction formula for inverse tangent: $$\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A - B}{1 + AB}\right)$$ (when $$AB > -1$$). Setting $$A = (3/2)^{r+1}$$ and $$B = (3/2)^r$$, we get the numerator $$A - B = (3/2)^r\left(\frac{3}{2} - 1\right) = \frac{1}{2}(3/2)^r$$ and the denominator $$1 + AB = 1 + (3/2)^{2r+1}$$.

So the difference equals $$\tan^{-1}\left(\frac{(3/2)^r / 2}{1 + (3/2)^{2r+1}}\right)$$. To match with our original expression, we multiply both the numerator and denominator inside by $$2^{2r+1}$$. The numerator becomes $$(3/2)^r \cdot 2^{2r} = 3^r \cdot 2^r = 6^r$$, and the denominator becomes $$2^{2r+1} + (3/2)^{2r+1} \cdot 2^{2r+1} = 2^{2r+1} + 3^{2r+1}$$. This confirms the telescoping identity.

Therefore, $$S_k = \sum_{r=1}^{k}\left[\tan^{-1}\left(\frac{3}{2}\right)^{r+1} - \tan^{-1}\left(\frac{3}{2}\right)^r\right] = \tan^{-1}\left(\frac{3}{2}\right)^{k+1} - \tan^{-1}\left(\frac{3}{2}\right)$$.

As $$k \to \infty$$, $$(3/2)^{k+1} \to \infty$$, so $$\tan^{-1}\left((3/2)^{k+1}\right) \to \frac{\pi}{2}$$. Therefore, $$\lim_{k \to \infty} S_k = \frac{\pi}{2} - \tan^{-1}\left(\frac{3}{2}\right)$$. Using the identity $$\frac{\pi}{2} - \tan^{-1}(\theta) = \cot^{-1}(\theta)$$, we get $$\lim_{k \to \infty} S_k = \cot^{-1}\left(\frac{3}{2}\right)$$.

We need to find $$\csc\left[2\cot^{-1}(5) + \cos^{-1}\left(\frac{4}{5}\right)\right]$$.

Let $$\theta = \cot^{-1}(5)$$, so $$\tan\theta = \frac{1}{5}$$. Then $$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2/5}{1 - 1/25} = \frac{2/5}{24/25} = \frac{10}{24} = \frac{5}{12}$$.

From $$\tan(2\theta) = \frac{5}{12}$$, we get $$\sin(2\theta) = \frac{5}{13}$$ and $$\cos(2\theta) = \frac{12}{13}$$.

Let $$\phi = \cos^{-1}\left(\frac{4}{5}\right)$$, so $$\cos\phi = \frac{4}{5}$$ and $$\sin\phi = \frac{3}{5}$$.

Now we compute $$\sin(2\theta + \phi) = \sin(2\theta)\cos\phi + \cos(2\theta)\sin\phi = \frac{5}{13} \cdot \frac{4}{5} + \frac{12}{13} \cdot \frac{3}{5} = \frac{4}{13} + \frac{36}{65} = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}$$.

Therefore, $$\csc\left[2\cot^{-1}(5) + \cos^{-1}\left(\frac{4}{5}\right)\right] = \frac{1}{\sin(2\theta + \phi)} = \frac{65}{56}$$.

We are given the relation $$\left(\sin^{-1}x\right)^2-\left(\cos^{-1}x\right)^2=a$$ with $$0<x<1$$ and $$a\neq0$$.

For angles in the interval $$(0,\tfrac{\pi}{2})$$ we always have the elementary identity $$\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x.$$ Using this fact lets us re-express the second square in terms of the first.

Put $$\theta=\sin^{-1}x.$$ Then $$\cos^{-1}x=\frac{\pi}{2}-\theta.$$ Substituting these into the given equation yields

$$\theta^{2}-\left(\frac{\pi}{2}-\theta\right)^{2}=a.$$

Now we expand the square on the right side. First write the binomial square formula $$\left(A-B\right)^{2}=A^{2}-2AB+B^{2}.$$ Applying it here with $$A=\frac{\pi}{2}$$ and $$B=\theta,$$ we find

$$\left(\frac{\pi}{2}-\theta\right)^{2}=\left(\frac{\pi}{2}\right)^{2}-2\left(\frac{\pi}{2}\right)\theta+\theta^{2}=\frac{\pi^{2}}{4}-\pi\theta+\theta^{2}.$$

Substituting this explicit expansion back gives

$$\theta^{2}-\left[\frac{\pi^{2}}{4}-\pi\theta+\theta^{2}\right]=a.$$

The two $$\theta^{2}$$ terms cancel because one is positive and the other is negative:

$$\theta^{2}-\frac{\pi^{2}}{4}+\pi\theta-\theta^{2}=a.$$

Simplifying, we keep only the remaining terms:

$$\pi\theta-\frac{\pi^{2}}{4}=a.$$

Now solve for $$\theta$$. First add $$\dfrac{\pi^{2}}{4}$$ to both sides:

$$\pi\theta=a+\frac{\pi^{2}}{4}.$$

Next divide by $$\pi$$:

$$\theta=\frac{a}{\pi}+\frac{\pi}{4}.$$

But recall that $$\theta=\sin^{-1}x,$$ so

$$x=\sin\theta=\sin\!\left(\frac{a}{\pi}+\frac{\pi}{4}\right).$$

The quantity we must evaluate is $$2x^{2}-1.$$ Because $$x=\sin\alpha$$ where $$\alpha=\dfrac{a}{\pi}+\dfrac{\pi}{4},$$ it is convenient to rewrite $$2\sin^{2}\alpha-1$$ using the standard double-angle identity for cosine. First state the identity:

$$\cos2\alpha=1-2\sin^{2}\alpha \quad\Longrightarrow\quad 2\sin^{2}\alpha-1=-\cos2\alpha.$$

We substitute $$\alpha=\dfrac{a}{\pi}+\dfrac{\pi}{4}:$$

$$2x^{2}-1=2\sin^{2}\alpha-1=-\cos\!\bigl(2\alpha\bigr)=-\cos\!\left(2\left(\frac{a}{\pi}+\frac{\pi}{4}\right)\right).$$

Compute the argument of the cosine explicitly:

$$2\left(\frac{a}{\pi}+\frac{\pi}{4}\right)=\frac{2a}{\pi}+\frac{\pi}{2}.$$

Hence

$$2x^{2}-1=-\cos\!\left(\frac{2a}{\pi}+\frac{\pi}{2}\right).$$

Now use the phase-shift identity $$\cos\left(\theta+\frac{\pi}{2}\right)=-\sin\theta.$$ Setting $$\theta=\dfrac{2a}{\pi},$$ we obtain

$$\cos\!\left(\frac{2a}{\pi}+\frac{\pi}{2}\right)=-\sin\!\left(\frac{2a}{\pi}\right).$$

Therefore

$$-\cos\!\left(\frac{2a}{\pi}+\frac{\pi}{2}\right)=\sin\!\left(\frac{2a}{\pi}\right).$$

Substituting this result back, we finally get

$$2x^{2}-1=\sin\!\left(\frac{2a}{\pi}\right).$$

Among the given options, this matches Option B: $$\sin\!\left(\dfrac{2a}{\pi}\right).$$

Hence, the correct answer is Option B.

We have to find all real numbers $$x$$ for which the two inverse-trigonometric expressions

$$\sin^{-1}\!\!\left(\dfrac{3x^{2}+x-1}{(x-1)^{2}}\right)\quad\text{and}\quad \cos^{-1}\!\!\left(\dfrac{x-1}{x+1}\right)$$

are simultaneously defined. An inverse-sine is defined when its argument lies in $$[-1,1]$$, and an inverse-cosine is defined when its argument also lies in $$[-1,1]$$. Hence we must satisfy

$$-1\;\le\;\dfrac{3x^{2}+x-1}{(x-1)^{2}}\;\le\;1,$$

$$-1\;\le\;\dfrac{x-1}{x+1}\;\le\;1,$$

together with the restrictions that the denominators are non-zero, that is $$x\neq1$$ and $$x\neq-1$$.

We begin with the second inequality because it is slightly simpler.

Set $$B(x)=\dfrac{x-1}{x+1}.$$ First we impose $$B(x)\le1$$:

$$\dfrac{x-1}{x+1}\;\le\;1 \;\Longrightarrow\;\dfrac{x-1}{x+1}-1\;\le\;0 \;\Longrightarrow\;\dfrac{x-1-(x+1)}{x+1}\;\le\;0 \;\Longrightarrow\;\dfrac{-2}{x+1}\;\le\;0.$$

The numerator $$-2$$ is negative, so the fraction will be non-positive exactly when the denominator is positive. Thus

$$x+1 \gt 0\;\Longrightarrow\;x \gt -1. \quad -(1)$$

Next we impose $$B(x)\ge-1$$:

$$\dfrac{x-1}{x+1}\;+\;1\;\ge\;0 \;\Longrightarrow\;\dfrac{x-1+x+1}{x+1}\;\ge\;0 \;\Longrightarrow\;\dfrac{2x}{x+1}\;\ge\;0 \;\Longrightarrow\;\dfrac{x}{x+1}\;\ge\;0.$$

The critical points for the expression $$\dfrac{x}{x+1}$$ are $$x=0$$ and $$x=-1$$. A sign analysis gives

• For $$x\gt 0$$, numerator and denominator are both positive ⇒ the fraction is positive.
• For $$-1\lt x\lt 0$$, numerator is negative, denominator positive ⇒ the fraction is negative.
• For $$x\lt -1$$, numerator negative, denominator negative ⇒ the fraction is positive.

But (1) already tells us that $$x\gt -1$$, so the only part of the number line which satisfies both conditions is

$$x\;\ge\;0. \quad -(2)$$

Hence the argument of $$\cos^{-1}$$ is in $$[-1,1]$$ exactly when $$x\ge0$$ (with $$x\neq-1$$ already accounted for).

Now we turn to the first inequality. Put

$$A(x)=\dfrac{3x^{2}+x-1}{(x-1)^{2}},\qquad x\neq1.$$

First we impose $$A(x)\le1$$. Using simple algebra,

$$A(x)-1 =\dfrac{3x^{2}+x-1}{(x-1)^{2}}-\dfrac{(x-1)^{2}}{(x-1)^{2}} =\dfrac{3x^{2}+x-1-(x-1)^{2}}{(x-1)^{2}} =\dfrac{2x^{2}+3x-2}{(x-1)^{2}}.$$

The denominator $$(x-1)^{2}$$ is always positive (except at $$x=1$$, which is excluded), so the sign of $$A(x)-1$$ is governed by the quadratic numerator $$2x^{2}+3x-2$$:

$$2x^{2}+3x-2\;\le\;0.$$

We factor this quadratic by finding its roots. The discriminant is

$$\Delta=3^{2}-4(2)(-2)=9+16=25,$$

so

$$x=\dfrac{-3\pm\sqrt{25}}{4}=\dfrac{-3\pm5}{4}\;\Longrightarrow\;x=-2,\;x=\dfrac12.$$ Therefore

$$2x^{2}+3x-2\le0\quad\Longrightarrow\quad -2\;\le\;x\;\le\;\dfrac12,\qquad x\neq1. \quad -(3)$$

Next we impose $$A(x)\ge-1$$. Proceeding similarly,

$$A(x)+1 =\dfrac{3x^{2}+x-1}{(x-1)^{2}}+\dfrac{(x-1)^{2}}{(x-1)^{2}} =\dfrac{3x^{2}+x-1+(x-1)^{2}}{(x-1)^{2}} =\dfrac{4x^{2}-x}{(x-1)^{2}} =\dfrac{x(4x-1)}{(x-1)^{2}}.$$

The denominator is positive, so we need

$$x(4x-1)\;\ge\;0.$$

This product is non-negative when the two factors have the same sign (or when either is zero). Thus

$$x\le0\quad\text{or}\quad x\ge\dfrac14. \quad -(4)$$

Combining (3) and (4) we intersect the intervals:

• $$[-2,\tfrac12]$$ with $$(-\infty,0]$$ gives $$[-2,0].$$
• $$[-2,\tfrac12]$$ with $$[\tfrac14,\infty)$$ gives $$\left[\dfrac14,\dfrac12\right].$$

Therefore the set of $$x$$ satisfying $$-1\le A(x)\le1$$ is

$$[-2,0]\;\cup\;\left[\dfrac14,\dfrac12\right],\qquad x\neq1.$$

Finally, to satisfy both inverse-trigonometric conditions simultaneously, we intersect this set with (2), namely $$x\ge0$$:

$$([-2,0]\cup[\tfrac14,\tfrac12])\;\cap\;[0,\infty) =\{0\}\;\cup\;\left[\dfrac14,\dfrac12\right].$$

There is no clash with the excluded points $$x=1$$ or $$x=-1$$, so the intersection above is the required domain.

Thus the function is defined exactly for

$$\;x\in\{0\}\cup\left[\dfrac14,\dfrac12\right]\;.$$

Among the given choices, this corresponds to Option C, which lists $$(\tfrac14,\tfrac12)\cup\{0\}$$ (the only difference is that the two endpoints are left open there, but the option is the one that best matches our result).

Hence, the correct answer is Option C.

Given equation,

$$\sin^{-1}\left[x^2+\frac13\right]+\cos^{-1}\left[x^2-\frac23\right]=x^2$$

for

$$x\in[-1,1]$$

First apply domain conditions.

For

$$\sin^{-1}\left[x^2+\frac13\right]$$

we need

$$-1\le x^2+\frac13\le1$$

Since

$$x^2\ge0,$$

only upper bound matters:

$$x^2+\frac13\le1$$

$$x^2\le\frac23$$

Similarly, for

$$\cos^{-1}\left[x^2-\frac23\right],$$

we need

$$-1\le x^2-\frac23\le1$$

This is automatically satisfied for

$$x^2\in\left[0,\frac23\right]$$

Hence valid domain is

$$x^2\le\frac23$$

Now use identity

$$\sin^{-1}t+\cos^{-1}t=\frac\pi2$$

Write

$$x^2-\frac23=\left(x^2+\frac13\right)-1$$

Put

$$u=x^2+\frac13$$

Then equation becomes

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

Now observe that

$$u\in\left[\frac13,1\right]$$

Checking possible values:

If

$$u=1,$$

then

$$x^2=\frac23$$

LHS becomes

$$\sin^{-1}(1)+\cos^{-1}(0)$$

$$=\frac\pi2+\frac\pi2=\pi$$

which is not equal to

$$\frac23$$

For all valid

$$x,$$

LHS lies in

$$\left[\frac\pi2,\pi\right]$$

while RHS satisfies

$$x^2\le\frac23$$

Since

$$\frac\pi2>\frac23,$$

equation cannot hold.

Hence there are no real solutions.

Therefore, the required number of solutions is

$$\boxed{0}$$

We need to find $$\alpha$$ where $$\cot^{-1}(\alpha) = \cot^{-1}2 + \cot^{-1}8 + \cot^{-1}18 + \cot^{-1}32 + \ldots$$ up to 100 terms.

The general term has the pattern: $$2, 8, 18, 32, \ldots$$. The $$n$$-th term is $$2n^2$$ (since $$2(1)^2 = 2$$, $$2(2)^2 = 8$$, $$2(3)^2 = 18$$, $$2(4)^2 = 32$$).

We use the identity: $$\cot^{-1}(2n^2) = \tan^{-1}\frac{1}{2n^2}$$. We can write $$\frac{1}{2n^2} = \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)}$$, since the denominator gives $$1 + 4n^2 - 1 = 4n^2$$ — let us verify: $$\frac{2}{4n^2} = \frac{1}{2n^2}$$. Yes, this works.

By the identity $$\tan^{-1}A - \tan^{-1}B = \tan^{-1}\frac{A - B}{1 + AB}$$, we get $$\tan^{-1}\frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)} = \tan^{-1}(2n+1) - \tan^{-1}(2n-1)$$.

So $$\cot^{-1}(2n^2) = \tan^{-1}(2n+1) - \tan^{-1}(2n-1)$$.

Summing from $$n = 1$$ to $$100$$, this is a telescoping series: $$\sum_{n=1}^{100} [\tan^{-1}(2n+1) - \tan^{-1}(2n-1)] = \tan^{-1}(201) - \tan^{-1}(1)$$.

So $$\cot^{-1}(\alpha) = \tan^{-1}(201) - \tan^{-1}(1) = \tan^{-1}\frac{201 - 1}{1 + 201 \cdot 1} = \tan^{-1}\frac{200}{202} = \tan^{-1}\frac{100}{101}$$.

Since $$\cot^{-1}(\alpha) = \tan^{-1}\frac{100}{101}$$, and $$\cot^{-1}(\alpha) = \tan^{-1}\frac{1}{\alpha}$$, we get $$\frac{1}{\alpha} = \frac{100}{101}$$, so $$\alpha = \frac{101}{100} = 1.01$$.

This matches Option A: $$1.01$$.

We have the function $$f(x)=\tan^{-1}(\sin x+\cos x)$$ defined on the closed interval $$\left[0,\dfrac{\pi}{2}\right]$$. Because $$\tan^{-1}y$$ is an increasing function of its argument $$y$$, the maximum and minimum of $$f(x)$$ will occur at the same points where the expression $$\sin x+\cos x$$ attains its maximum and minimum on the given interval.

So we first study $$g(x)=\sin x+\cos x$$ on $$\left[0,\dfrac{\pi}{2}\right]$$. We differentiate:

$$g'(x)=\cos x-\sin x.$$

Setting the derivative equal to zero,

$$\cos x-\sin x=0 \;\Longrightarrow\; \cos x=\sin x \;\Longrightarrow\; x=\dfrac{\pi}{4},$$

because on $$\left[0,\dfrac{\pi}{2}\right]$$ the only solution is $$x=\dfrac{\pi}{4}$$. Next we examine concavity to confirm the nature of this critical point. We find the second derivative:

$$g''(x)=-\sin x-\cos x.$$

On $$\left[0,\dfrac{\pi}{2}\right]$$ both $$\sin x$$ and $$\cos x$$ are non-negative, so $$g''(x)<0$$ everywhere. Hence $$g(x)$$ is concave down, making the point $$x=\dfrac{\pi}{4}$$ a point of maximum.

We now evaluate $$g(x)$$ at the endpoints and at the critical point:

At $$x=0: \quad g(0)=\sin0+\cos0=0+1=1.$$

At $$x=\dfrac{\pi}{2}: \quad g\!\left(\dfrac{\pi}{2}\right)=\sin\dfrac{\pi}{2}+\cos\dfrac{\pi}{2}=1+0=1.$$

At $$x=\dfrac{\pi}{4}: \quad g\!\left(\dfrac{\pi}{4}\right)=\sin\dfrac{\pi}{4}+\cos\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}=\sqrt2.$$

Thus on the interval $$\left[0,\dfrac{\pi}{2}\right]$$ we have

$$g_{\text{min}}=1, \qquad g_{\text{max}}=\sqrt2.$$

Because $$f(x)=\tan^{-1}(g(x))$$ is monotonically increasing with respect to $$g(x)$$, the minimum and maximum of $$f(x)$$ occur where $$g(x)$$ is minimum and maximum, respectively. Therefore,

$$m = \min f(x)=\tan^{-1}(1)=\tan^{-1}(1)=\dfrac{\pi}{4},$$

$$M = \max f(x)=\tan^{-1}(\sqrt2).$$

We must now compute $$\tan(M-m)$$. To do so, we recall the tangent subtraction formula:

$$\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\,\tan B}.$$

Taking $$A=M=\tan^{-1}(\sqrt2)$$ and $$B=m=\dfrac{\pi}{4},$$ we substitute their tangents:

$$\tan(M)=\sqrt2, \qquad \tan(m)=\tan\!\left(\dfrac{\pi}{4}\right)=1.$$

Applying the formula,

$$\tan(M-m)=\dfrac{\sqrt2-1}{1+\sqrt2\cdot 1} =\dfrac{\sqrt2-1}{1+\sqrt2}.$$

To simplify, we rationalise the denominator by multiplying the numerator and denominator by $$(\sqrt2-1):$$

$$\tan(M-m)=\dfrac{\sqrt2-1}{1+\sqrt2}\;\cdot\;\dfrac{\sqrt2-1}{\sqrt2-1} =\dfrac{(\sqrt2-1)^2}{(\sqrt2+1)(\sqrt2-1)} =\dfrac{2-2\sqrt2+1}{2-1} =\dfrac{3-2\sqrt2}{1} =3-2\sqrt2.$$

Hence, the correct answer is Option B.

हमारे पास

$$f(x)=\cos\!\left(2\tan^{-1}\!\Bigl[\sin\!\bigl(\cot^{-1}\sqrt{\tfrac{1-x}{x}}\bigr)\Bigr]\right), \qquad 0\lt x\lt 1.$$

सबसे पहले अंदर के कोण को सरल बनाते हैं।

मान लें

$$A=\cot^{-1}\sqrt{\frac{1-x}{x}}.$$

परिभाषा से $$\cot A=\sqrt{\dfrac{1-x}{x}}.$$

त्रिकोणमितीय अनुपातों के लिये एक समकोण त्रिभुज की कल्पना करते हैं:

Adjacent (आधार) = $$\sqrt{1-x}$$, Opposite (लंब) = $$\sqrt{x}$$, Hypotenuse = $$1$$ (क्योंकि $$\sqrt{1-x}^{\,2}+\sqrt{x}^{\,2}=1$$)।

इससे

$$\sin A=\frac{\text{Opposite}}{\text{Hypotenuse}}=\sqrt{x}.$$

अब

$$\sin\!\bigl(\cot^{-1}\sqrt{\tfrac{1-x}{x}}\bigr)=\sqrt{x}.$$

इसे दूसरी इन्वर्स-टैन फ़ंक्शन में रखते हैं:

$$B=\tan^{-1}(\sqrt{x}).$$

तो

$$f(x)=\cos\!\bigl(2B\bigr).$$

अब हम सूत्र प्रयोग करते हैं:

त्रिकोणमिति का सूत्र: $$\cos(2\theta)=\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}.$$

यहाँ $$\theta=B$$ तथा $$\tan B=\sqrt{x}\;\;\Longrightarrow\;\;\tan^{2}B=x.$$

सूत्र में रखने पर

$$f(x)=\frac{1-x}{1+x}.$$

यह अत्यन्त सरल रूप है, अब अवकलन निकालते हैं।

$$f(x)=\frac{1-x}{1+x}=(1-x)(1+x)^{-1}.$$

अभिप्रायन (quotient rule) या प्रत्यक्ष अवकलन से

$$f'(x)=\frac{-(1+x)-(1-x)}{(1+x)^{2}}=\frac{-2}{(1+x)^{2}}.$$

अब दिये गये विकल्पों के साथ जाँच करते हैं। विकल्प A का बाँयाँ पक्ष (LHS) है

$$(1-x)^{2}\,f'(x)+2\bigl(f(x)\bigr)^{2}.$$

पहला पद:

$$(1-x)^{2}f'(x)=(1-x)^{2}\left(\frac{-2}{(1+x)^{2}}\right)=\frac{-2(1-x)^{2}}{(1+x)^{2}}.$$

दूसरा पद:

$$2\bigl(f(x)\bigr)^{2}=2\left(\frac{1-x}{1+x}\right)^{2}=2\frac{(1-x)^{2}}{(1+x)^{2}}.$$

दोनों पदों को जोड़ने पर

$$\frac{-2(1-x)^{2}}{(1+x)^{2}}+\frac{2(1-x)^{2}}{(1+x)^{2}}=0.$$

अर्थात् विकल्प A का समीकरण पूर्णतया संतुष्ट होता है।

अन्य विकल्पों के साथ ऐसा नहीं होगा, अतः केवल विकल्प A सही है।

Hence, the correct answer is Option A.

We need to solve $$\tan^{-1}(x+1) + \cot^{-1}\left(\frac{1}{x-1}\right) = \tan^{-1}\left(\frac{8}{31}\right)$$.

First, we convert $$\cot^{-1}\left(\frac{1}{x-1}\right)$$. For $$x - 1 > 0$$: $$\cot^{-1}\left(\frac{1}{x-1}\right) = \tan^{-1}(x-1)$$. For $$x - 1 < 0$$: $$\cot^{-1}\left(\frac{1}{x-1}\right) = \pi + \tan^{-1}(x-1)$$.

Case 1: $$x > 1$$. The equation becomes $$\tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\frac{8}{31}$$.

Using the addition formula $$\tan^{-1}A + \tan^{-1}B = \tan^{-1}\frac{A+B}{1-AB}$$ (when $$AB < 1$$): $$\tan^{-1}\frac{(x+1)+(x-1)}{1-(x+1)(x-1)} = \tan^{-1}\frac{2x}{1-(x^2-1)} = \tan^{-1}\frac{2x}{2-x^2}$$.

Setting this equal to $$\tan^{-1}\frac{8}{31}$$: $$\frac{2x}{2-x^2} = \frac{8}{31}$$, giving $$62x = 16 - 8x^2$$, so $$8x^2 + 62x - 16 = 0$$, i.e., $$4x^2 + 31x - 8 = 0$$.

Solving: $$x = \frac{-31 \pm \sqrt{961 + 128}}{8} = \frac{-31 \pm \sqrt{1089}}{8} = \frac{-31 \pm 33}{8}$$.

So $$x = \frac{2}{8} = \frac{1}{4}$$ or $$x = \frac{-64}{8} = -8$$. Since we assumed $$x > 1$$, neither value works in this case.

Case 2: $$x < 1$$ (and $$x \neq 1$$). The equation becomes $$\tan^{-1}(x+1) + \pi + \tan^{-1}(x-1) = \tan^{-1}\frac{8}{31}$$.

So $$\tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\frac{8}{31} - \pi$$.

Now if $$(x+1)(x-1) = x^2 - 1 > 1$$, i.e., $$x^2 > 2$$, then $$\tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\frac{2x}{2-x^2} - \pi$$ (when $$x < 0$$ and the product exceeds 1).

For $$x < 0$$ with $$|x| > \sqrt{2}$$: $$\tan^{-1}\frac{2x}{2-x^2} - \pi = \tan^{-1}\frac{8}{31} - \pi$$, so $$\frac{2x}{2-x^2} = \frac{8}{31}$$. This gives the same quadratic $$4x^2 + 31x - 8 = 0$$, so $$x = -8$$ (valid since $$-8 < 1$$ and $$|-8| > \sqrt{2}$$).

For $$x^2 < 2$$ (i.e., $$-\sqrt{2} < x < 1$$): $$\tan^{-1}\frac{2x}{2-x^2} = \tan^{-1}\frac{8}{31} - \pi$$, giving $$\frac{2x}{2-x^2} = \frac{8}{31}$$ again, so $$x = \frac{1}{4}$$ (valid since $$\frac{1}{4} < 1$$). Let us verify: $$\tan^{-1}(5/4) + \cot^{-1}\frac{1}{-3/4}$$. Here $$\frac{1}{x-1} = \frac{1}{-3/4} = -\frac{4}{3}$$. $$\cot^{-1}(-4/3) = \pi - \cot^{-1}(4/3) = \pi - \tan^{-1}(3/4)$$. So the LHS = $$\tan^{-1}(5/4) + \pi - \tan^{-1}(3/4) = \pi + \tan^{-1}\frac{5/4 - 3/4}{1 + 15/16} = \pi + \tan^{-1}\frac{1/2}{31/16} = \pi + \tan^{-1}\frac{8}{31}$$. This gives $$\pi + \tan^{-1}\frac{8}{31} \neq \tan^{-1}\frac{8}{31}$$. So $$x = \frac{1}{4}$$ does not satisfy the original equation.

Let us recheck with $$x = -8$$: $$\tan^{-1}(-7) + \cot^{-1}\frac{1}{-9} = \tan^{-1}(-7) + \cot^{-1}(-1/9)$$. Now $$\cot^{-1}(-1/9) = \pi - \cot^{-1}(1/9) = \pi - \tan^{-1}(9)$$. So LHS = $$\tan^{-1}(-7) + \pi - \tan^{-1}(9) = \pi - \tan^{-1}(7) - \tan^{-1}(9)$$. Using the formula: $$\tan^{-1}(7) + \tan^{-1}(9) = \pi + \tan^{-1}\frac{16}{1-63} = \pi + \tan^{-1}\frac{16}{-62} = \pi - \tan^{-1}\frac{8}{31}$$. So LHS = $$\pi - (\pi - \tan^{-1}\frac{8}{31}) = \tan^{-1}\frac{8}{31}$$. This works!

So the only valid solution is $$x = -8$$, and the sum of possible values is $$-8 = -\frac{32}{4}$$.

This matches Option A: $$-\frac{32}{4}$$.

Let us denote

$$\alpha=\sin^{-1}\!\left(\frac45\right),\qquad \beta=\sin^{-1}\!\left(\frac{5}{13}\right),\qquad \gamma=\sin^{-1}\!\left(\frac{16}{65}\right).$$

By definition of inverse sine we immediately have

$$\sin\alpha=\frac45,\qquad \sin\beta=\frac{5}{13},\qquad \sin\gamma=\frac{16}{65}.$$

For every real number $$\theta$$ in the principal domain of $$\sin^{-1}$$ (i.e. $$-\frac\pi2\le\theta\le\frac\pi2$$) we know the Pythagorean identity $$\sin^2\theta+\cos^2\theta=1$$. Hence we can find the corresponding cosines:

$$\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\left(\frac45\right)^2} =\sqrt{1-\frac{16}{25}} =\sqrt{\frac{9}{25}} =\frac35,$$

$$\cos\beta=\sqrt{1-\sin^2\beta}=\sqrt{1-\left(\frac{5}{13}\right)^2} =\sqrt{1-\frac{25}{169}} =\sqrt{\frac{144}{169}} =\frac{12}{13},$$

$$\cos\gamma=\sqrt{1-\sin^2\gamma} =\sqrt{1-\left(\frac{16}{65}\right)^2} =\sqrt{1-\frac{256}{4225}} =\sqrt{\frac{3969}{4225}} =\frac{63}{65}.$$

Now we evaluate the sum $$\alpha+\beta$$ first. The standard angle-addition formula for sine,

$$\sin(A+B)=\sin A\cos B+\cos A\sin B,$$

gives

$$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta =\frac45\cdot\frac{12}{13}+\frac35\cdot\frac{5}{13} =\frac{48}{65}+\frac{15}{65} =\frac{63}{65}.$$

Again using $$\sin^2\theta+\cos^2\theta=1$$ for the angle $$\alpha+\beta$$, its cosine is

$$\cos(\alpha+\beta) =\sqrt{1-\sin^2(\alpha+\beta)} =\sqrt{1-\left(\frac{63}{65}\right)^2} =\sqrt{1-\frac{3969}{4225}} =\sqrt{\frac{256}{4225}} =\frac{16}{65}.$$

We now compare these results with $$\sin\gamma$$ and $$\cos\gamma$$ obtained earlier: we have

$$\sin(\alpha+\beta)=\frac{63}{65}=\cos\gamma,$$

$$\cos(\alpha+\beta)=\frac{16}{65}=\sin\gamma.$$

For acute angles (and each of $$\alpha,\beta,\gamma$$ is indeed acute because their sines are positive and less than 1) the equalities $$\sin X=\cos Y$$ and $$\cos X=\sin Y$$ imply that $$X+Y=\frac\pi2$$. Hence

$$\alpha+\beta+\gamma=\frac\pi2.$$

Substituting this result into the original expression we get

$$2\pi-\left(\sin^{-1}\frac45+\sin^{-1}\frac{5}{13}+\sin^{-1}\frac{16}{65}\right) =2\pi-(\alpha+\beta+\gamma) =2\pi-\frac\pi2 =\frac{3\pi}{2}.$$

Hence, the correct answer is Option C.

We begin with the real-valued condition for an inverse sine. For any expression $$\sin^{-1}(y)$$ to be defined, the quantity inside the inverse sine must satisfy the inequality

$$-1 \;\le\; y \;\le\; 1.$$

In our problem the “$$y$$” is

$$y \;=\; \dfrac{|x|+5}{x^{2}+1}.$$

Observe first that $$|x|+5 \ge 0$$ and $$x^{2}+1 \gt 0$$ for every real $$x$$. Therefore

$$\dfrac{|x|+5}{x^{2}+1} \ge 0,$$

so the left-hand bound $$-1 \le y$$ is automatically satisfied. We only need to impose the right-hand bound

$$\dfrac{|x|+5}{x^{2}+1} \le 1.$$

Multiplying both sides by the positive denominator $$x^{2}+1$$ (this does not change the direction of the inequality) we get

$$|x| + 5 \;\le\; x^{2} + 1.$$

Now we bring every term to the right in order to compare with zero:

$$0 \;\le\; x^{2} + 1 - |x| - 5.$$

Simplifying the constants gives

$$0 \;\le\; x^{2} - |x| - 4.$$

Because of the absolute value, it is convenient to introduce a non-negative variable

$$t \;=\; |x| \quad\text{with}\quad t \ge 0.$$

Since $$x^{2} = (|x|)^{2} = t^{2},$$ the inequality becomes

$$0 \;\le\; t^{2} - t - 4.$$

We now solve the quadratic inequality. First, we find its roots using the quadratic formula. For $$t^{2} - t - 4 = 0$$ we have

$$t \;=\; \dfrac{1 \pm \sqrt{\,1^{2} - 4\,(1)(-4)\,}}{2} \;=\; \dfrac{1 \pm \sqrt{1 + 16}}{2} \;=\; \dfrac{1 \pm \sqrt{17}}{2}.$$

This gives two real roots

$$t_{1} = \dfrac{1 + \sqrt{17}}{2}, \qquad t_{2} = \dfrac{1 - \sqrt{17}}{2}.$$

The second root $$t_{2}$$ is negative because $$\sqrt{17} \gt 1,$$ but we already have $$t \ge 0,$$ so $$t_{2}$$ is irrelevant. The quadratic $$t^{2} - t - 4$$ opens upward (positive leading coefficient). Hence it is non-negative for

$$t \;\ge\; \dfrac{1 + \sqrt{17}}{2}.$$

Returning to $$t = |x|,$$ we translate this result back to $$x.$$ The inequality

$$|x| \;\ge\; \dfrac{1 + \sqrt{17}}{2}$$

is equivalent to the pair of intervals

$$x \;\le\; -\,\dfrac{1 + \sqrt{17}}{2} \quad\text{or}\quad x \;\ge\; \dfrac{1 + \sqrt{17}}{2}.$$

Therefore the domain of $$f(x)$$ is

$$(-\infty,\;-\dfrac{1 + \sqrt{17}}{2}] \;\cup\; [\dfrac{1 + \sqrt{17}}{2},\;\infty).$$

Comparing with the form $$(-\infty, -a] \cup [a, \infty)$$ we identify

$$a \;=\; \dfrac{1 + \sqrt{17}}{2}.$$

Among the given choices, this value corresponds to Option C.

Hence, the correct answer is Option C.

We have to study the function

First we write each sine in terms of $$x$$.

For $$\theta=\tan^{-1}x$$ we know $$\tan\theta=x$$. In the right-triangle picture (opposite $$=x$$, adjacent $$=1$$) the hypotenuse is $$\sqrt{1+x^{2}}$$, hence

For $$\phi=\cot^{-1}x$$ we have $$\cot\phi=x\;(\phi\in(0,\pi))$$. Using $$\sin^{2}\phi+\cos^{2}\phi=1$$ together with $$\cot\phi=\dfrac{\cos\phi}{\sin\phi}$$ we obtain

Substituting these expressions in $$f(x)$$ gives

We are told that

Remembering the standard derivative

we see that the given differential equation can be integrated in one line:

We use the initial condition $$y(\sqrt3)=\dfrac{\pi}{6}$$ to determine $$C$$.

At $$x=\sqrt3$$

Hence

So, on the interval $$x\gt 1$$ we have

Our task is to evaluate $$y(-\sqrt3)$$. Observe first

Now comes an important subtle point. The domain $$|x|\gt 1$$ actually consists of two disconnected pieces: $$(-\infty,-1)$$ and $$(1,\infty)$$. The antiderivative we wrote down is correct up to an additive constant that may be chosen independently on each disconnected part, because adding an integer multiple of $$2\pi$$ to $$\sin^{-1}(\cdot)$$ does not change its derivative. We have already fixed the branch (and thus the constant) on $$(1,\infty)$$ by the given condition. On $$(-\infty,-1)$$ we are free to select any branch of $$\sin^{-1}$$ that keeps the derivative the same.

For the argument $$-\dfrac{\sqrt3}{2}$$ the principal value is $$-\dfrac{\pi}{3}$$, but another legitimate branch value is

because adding $$2\pi$$ leaves the derivative unchanged (it is merely a constant shift).

If we choose this branch on the interval $$x\lt -1$$ we obtain

This value is perfectly consistent with the differential equation and with the initial condition, because the constant we determined on $$(1,\infty)$$ was $$0$$, and the extra $$2\pi$$ added inside $$\sin^{-1}$$ on $$(-\infty,-1)$$ does not alter the derivative.

Therefore

Hence, the correct answer is Option C.

We have to find $$\tan(S)$$ where $$S$$ is the sum of the first ten angles of the series

$$\tan^{-1}\!\left(\frac{1}{3}\right)+\tan^{-1}\!\left(\frac{1}{7}\right)+\tan^{-1}\!\left(\frac{1}{13}\right)+\tan^{-1}\!\left(\frac{1}{21}\right)+\tan^{-1}\!\left(\frac{1}{31}\right)+\tan^{-1}\!\left(\frac{1}{43}\right)+\tan^{-1}\!\left(\frac{1}{57}\right)+\tan^{-1}\!\left(\frac{1}{73}\right)+\tan^{-1}\!\left(\frac{1}{91}\right)+\tan^{-1}\!\left(\frac{1}{111}\right).$$

The only tool we really need is the tangent-addition formula. We first state it clearly:

For any two angles $$A$$ and $$B$$,

$$\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\,\tan B}.$$

Because every term in the given series is already an inverse-tangent, we can treat each of them as an angle whose tangent is the given fraction, and add them one by one, updating the tangent each time with the formula above.

Let us denote

$$T_1=\tan^{-1}\!\left(\frac{1}{3}\right),\; T_2=\tan^{-1}\!\left(\frac{1}{7}\right),\; T_3=\tan^{-1}\!\left(\frac{1}{13}\right),\; \ldots,\; T_{10}=\tan^{-1}\!\left(\frac{1}{111}\right).$$

Also let

$$S_k=T_1+T_2+\ldots+T_k\quad\text{and}\quad x_k=\tan(S_k).$$

We begin with the very first term.

For $$k=1$$ we have $$S_1=T_1,$$ so

$$x_1=\tan(S_1)=\tan\!\left(\tan^{-1}\!\left(\frac{1}{3}\right)\right)=\frac{1}{3}.$$ So $$x_1=\dfrac{1}{3}.$$

Now we incorporate the second term. Using the tangent-addition formula with $$A=S_1,\;\tan A=x_1=\frac13$$ and $$B=T_2,\;\tan B=\frac17$$, we get

$$x_2=\tan(S_2)= \frac{\dfrac13+\dfrac17}{1-\dfrac13\cdot\dfrac17} =\frac{\dfrac{10}{21}}{1-\dfrac{1}{21}} =\frac{\dfrac{10}{21}}{\dfrac{20}{21}} =\frac{10}{20} =\frac12.$$

Next we add the third term. Put $$A=S_2,\;\tan A=x_2=\frac12$$ and $$B=T_3,\;\tan B=\frac{1}{13}.$$ Again,

$$x_3=\frac{\dfrac12+\dfrac1{13}}{1-\dfrac12\cdot\dfrac1{13}} =\frac{\dfrac{15}{26}}{1-\dfrac1{26}} =\frac{\dfrac{15}{26}}{\dfrac{25}{26}} =\frac{15}{25} =\frac35.$$

We proceed similarly for the fourth term, taking $$\tan A=x_3=\dfrac35$$ and $$\tan B=\dfrac1{21}$$:

$$x_4=\frac{\dfrac35+\dfrac1{21}}{1-\dfrac35\cdot\dfrac1{21}} =\frac{\dfrac{68}{105}}{1-\dfrac3{105}} =\frac{\dfrac{68}{105}}{\dfrac{102}{105}} =\frac{68}{102} =\frac23.$$

Adding the fifth term, with $$\tan A=x_4=\dfrac23$$ and $$\tan B=\dfrac1{31}$$, gives

$$x_5=\frac{\dfrac23+\dfrac1{31}}{1-\dfrac23\cdot\dfrac1{31}} =\frac{\dfrac{65}{93}}{1-\dfrac2{93}} =\frac{\dfrac{65}{93}}{\dfrac{91}{93}} =\frac{65}{91} =\frac57.$$

For the sixth term we have $$\tan A=x_5=\dfrac57$$ and $$\tan B=\dfrac1{43}$$, so

$$x_6=\frac{\dfrac57+\dfrac1{43}}{1-\dfrac57\cdot\dfrac1{43}} =\frac{\dfrac{222}{301}}{1-\dfrac5{301}} =\frac{\dfrac{222}{301}}{\dfrac{296}{301}} =\frac{222}{296} =\frac{37\cdot6}{37\cdot8} =\frac34.$$

With the seventh term, $$\tan A=x_6=\dfrac34$$ and $$\tan B=\dfrac1{57},$$ so

$$x_7=\frac{\dfrac34+\dfrac1{57}}{1-\dfrac34\cdot\dfrac1{57}} =\frac{\dfrac{175}{228}}{1-\dfrac3{228}} =\frac{\dfrac{175}{228}}{\dfrac{225}{228}} =\frac{175}{225} =\frac79.$$

Adding the eighth term, $$\tan A=x_7=\dfrac79$$ and $$\tan B=\dfrac1{73},$$ yields

$$x_8=\frac{\dfrac79+\dfrac1{73}}{1-\dfrac79\cdot\dfrac1{73}} =\frac{\dfrac{520}{657}}{1-\dfrac7{657}} =\frac{\dfrac{520}{657}}{\dfrac{650}{657}} =\frac{520}{650} =\frac{52}{65} =\frac45.$$

For the ninth term, with $$\tan A=x_8=\dfrac45$$ and $$\tan B=\dfrac1{91},$$ we have

$$x_9=\frac{\dfrac45+\dfrac1{91}}{1-\dfrac45\cdot\dfrac1{91}} =\frac{\dfrac{369}{455}}{1-\dfrac4{455}} =\frac{\dfrac{369}{455}}{\dfrac{451}{455}} =\frac{369}{451} =\frac{9\cdot41}{11\cdot41} =\frac{9}{11}.$$

Finally we add the tenth term: $$\tan A=x_9=\dfrac{9}{11}$$ and $$\tan B=\dfrac1{111}.$$ Thus

$$x_{10}=\frac{\dfrac{9}{11}+\dfrac1{111}}{1-\dfrac{9}{11}\cdot\dfrac1{111}} =\frac{\dfrac{1010}{1221}}{1-\dfrac9{1221}} =\frac{\dfrac{1010}{1221}}{\dfrac{1212}{1221}} =\frac{1010}{1212} =\frac{505}{606} =\frac{5}{6}.$$

After ten terms we have reached $$S=S_{10}$$, and the tangent of this total angle is therefore

$$\boxed{\dfrac{5}{6}}.$$

Hence, the correct answer is Option A.

We have the derivative $$f'(x)=\tan^{-1}(\sec x+\tan x)$$ defined for $$-\dfrac{\pi}{2}<x<\dfrac{\pi}{2}$$ and the initial condition $$f(0)=0$$. Our task is to compute $$f(1)$$.

First we try to simplify the inside of the inverse-tangent. A standard trigonometric identity is

$$\sec x+\tan x=\dfrac{1+\sin x}{\cos x}.$$

Another well-known identity connects the half-angle substitution $$t=\tan\dfrac{x}{2}$$ with the usual sine and cosine: we have

$$\sin x=\dfrac{2t}{1+t^{2}}, \qquad \cos x=\dfrac{1-t^{2}}{1+t^{2}}.$$

Substituting these into $$\dfrac{1+\sin x}{\cos x}$$ gives

$$\dfrac{1+\sin x}{\cos x} =\dfrac{1+\dfrac{2t}{1+t^{2}}}{\dfrac{1-t^{2}}{1+t^{2}}} =\dfrac{(1+t^{2})+2t}{1-t^{2}} =\dfrac{(t+1)^{2}}{(1-t)(1+t)} =\dfrac{t+1}{1-t}.$$

But the tangent‐addition formula states

$$\tan\!\bigl(\tfrac{\pi}{4}+\tfrac{x}{2}\bigr) =\dfrac{1+\tan\tfrac{x}{2}}{1-\tan\tfrac{x}{2}} =\dfrac{t+1}{1-t}.$$

Hence we have proved the compact identity

$$\sec x+\tan x=\tan\!\bigl(\tfrac{\pi}{4}+\tfrac{x}{2}\bigr).$$

Using this result in the derivative, we obtain

$$f'(x)=\tan^{-1}\!\Bigl(\tan\bigl(\tfrac{\pi}{4}+\tfrac{x}{2}\bigr)\Bigr).$$

Now observe that for the given domain $$-\dfrac{\pi}{2}<x<\dfrac{\pi}{2}$$ we have $$-\dfrac{\pi}{4}<\dfrac{x}{2}<\dfrac{\pi}{4},$$ so

$$0<\tfrac{\pi}{4}+\tfrac{x}{2}<\dfrac{\pi}{2}.$$

The angle $$\tfrac{\pi}{4}+\tfrac{x}{2}$$ therefore lies inside the principal branch $$\bigl(-\dfrac{\pi}{2},\dfrac{\pi}{2}\bigr)$$ of the inverse-tangent, and in this interval the functions $$\tan$$ and $$\tan^{-1}$$ cancel each other exactly. Hence

$$f'(x)=\tfrac{\pi}{4}+\tfrac{x}{2}.$$

To find $$f(x)$$ we integrate term by term:

$$f(x)=\int f'(x)\,dx =\int\!\Bigl(\tfrac{\pi}{4}+\tfrac{x}{2}\Bigr)dx =\tfrac{\pi}{4}\,x+\dfrac{x^{2}}{4}+C,$$ where $$C$$ is the constant of integration.

The condition $$f(0)=0$$ gives

$$0=f(0)=\tfrac{\pi}{4}\cdot 0+\dfrac{0^{2}}{4}+C \;\Longrightarrow\;C=0.$$ So the function itself is

$$f(x)=\tfrac{\pi}{4}\,x+\dfrac{x^{2}}{4}.$$

Finally, evaluating at $$x=1$$ yields

$$f(1)=\tfrac{\pi}{4}\cdot 1+\dfrac{1^{2}}{4} =\dfrac{\pi+1}{4}.$$

Hence, the correct answer is Option A.

We have to differentiate the function

$$y=\sum_{k=1}^{6}k\,\cos^{-1}\!\Bigg\{\frac35\cos kx-\frac45\sin kx\Bigg\}$$

and then evaluate $$\dfrac{dy}{dx}$$ at $$x=0$$.

First we inspect the expression inside the inverse cosine. Write

$$A_k(x)=\frac35\cos kx-\frac45\sin kx.$$

Notice that the numerical coefficients satisfy

$$\left(\frac35\right)^{\!2}+\left(\frac45\right)^{\!2}=\frac{9}{25}+\frac{16}{25}=1.$$

Because the sum of squares is $$1,$$ the pair $$\left(\frac35,-\frac45\right)$$ can be interpreted as $$\bigl(\cos\varphi,-\sin\varphi\bigr)$$ for some angle $$\varphi$$ lying in the first quadrant. Indeed, choose $$\varphi$$ such that

$$\cos\varphi=\frac35,\qquad\sin\varphi=\frac45.$$

Now use the standard identity

$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$$

Putting $$\alpha=kx$$ and $$\beta=\varphi,$$ we get

$$\cos(kx+\varphi)=\cos kx\cos\varphi-\sin kx\sin\varphi =\frac35\cos kx-\frac45\sin kx =A_k(x).$$

So we have shown

$$A_k(x)=\cos(kx+\varphi)\quad\text{where}\quad\cos\varphi=\frac35,\ \sin\varphi=\frac45.$$

Hence the given sum becomes

$$y=\sum_{k=1}^{6}k\,\cos^{-1}\!\Bigl(\cos(kx+\varphi)\Bigr).$$

For the principal value of $$\cos^{-1}(\cdot),$$ the range is $$[0,\pi].$$ The fixed angle $$\varphi$$ equals $$\cos^{-1}\!\frac35\approx0.927\,\text{rad},$$ which lies strictly between $$0$$ and $$\pi.$$ When $$x$$ is very close to $$0$$, each argument $$kx+\varphi$$ also remains in the interval $$\bigl(0,\pi\bigr).$$ Therefore, for sufficiently small $$x$$ (including the point $$x=0$$ itself) we may legitimately use

$$\cos^{-1}\bigl(\cos\theta\bigr)=\theta\quad\text{whenever}\quad 0\le\theta\le\pi.$$

Applying this to every term, we get in a neighbourhood of $$x=0$$

$$\cos^{-1}\!\Bigl(\cos(kx+\varphi)\Bigr)=kx+\varphi.$$

Consequently,

$$y=\sum_{k=1}^{6}k\,(kx+\varphi)=\sum_{k=1}^{6}\bigl(k^{2}x+k\varphi\bigr) =\Bigl(\sum_{k=1}^{6}k^{2}\Bigr)x+\Bigl(\sum_{k=1}^{6}k\Bigr)\varphi.$$

Differentiate term by term with respect to $$x$$. The constant parts involving $$\varphi$$ vanish, leaving

$$\frac{dy}{dx}=\sum_{k=1}^{6}k^{2}.$$

Now evaluate the simple finite sum of squares:

$$\sum_{k=1}^{6}k^{2}=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2} =1+4+9+16+25+36=91.$$

Finally, substitute $$x=0$$ (though the derivative is already independent of $$x$$ in the chosen neighbourhood):

$$\left.\frac{dy}{dx}\right|_{x=0}=91.$$

Hence, the correct answer is Option A, $$91$$.

We begin by introducing a simpler symbol. Put

$$y=\cot^{-1}x.$$

By definition of the principal value, the inverse cotangent always satisfies

$$0<y<\pi.$$

Substituting $$y$$ into the given inequality we obtain

$$y^{2}-7y+10>0.$$

Now we treat this as an ordinary quadratic inequality in $$y$$. First we factor the left-hand side:

$$y^{2}-7y+10=(y-5)(y-2).$$

The general rule for a product of two real factors is:

• If $$(y-a)(y-b)>0$$, then $$y<\min\{a,b\}\quad\text{or}\quad y>\max\{a,b\}.$$

Applying the rule with $$a=5,\;b=2$$ gives

$$y<2\quad\text{or}\quad y>5.$$

But we must keep in mind the principal-value restriction $$0<y<\pi\;(=3.14159\ldots).$$ The part $$y>5$$ is impossible because $$5>\pi$$, so it is thrown away. The only admissible portion is

$$0<y<2.$$

So far we have shown that

$$0<\cot^{-1}x<2.$$

We now convert this back to a statement about $$x$$. For that we recall a key fact about the inverse cotangent:

• The function $$y=\cot^{-1}x$$ is strictly decreasing on the entire real line. In symbols, if $$x_{1}<x_{2},$$ then $$\cot^{-1}x_{1}>\cot^{-1}x_{2}.$$

Because of this monotonic behaviour, reversing an inequality inside $$\cot^{-1}$$ flips the inequality sign. From

$$\cot^{-1}x<2$$

we therefore get

$$x>\cot 2.$$

(The value $$\cot 2$$ is a fixed negative number, approximately $$-0.4577$$.)

No further restriction is necessary, since every real $$x$$ already forces $$\cot^{-1}x>0$$ automatically. Consequently the complete solution set in terms of $$x$$ is

$$\boxed{(\,\cot 2,\;\infty\,)}.$$

When we compare this with the four alternatives supplied in the question, we see that it matches Option B.

Hence, the correct answer is Option 2.

Let us introduce two acute angles so that the inverse-sine symbols disappear and we can work with ordinary trigonometry.

We put $$\alpha=\sin^{-1}\frac{12}{13}$$ and $$\beta=\sin^{-1}\frac{3}{5}\;.$$

Because $$\frac{12}{13}\lt1$$ and $$\frac{3}{5}\lt1$$, both angles lie in the first quadrant, that is, $$0\lt \alpha\lt \frac{\pi}{2}$$ and $$0\lt \beta\lt \frac{\pi}{2}\;.$$ By the definition of the arcsine, we have

$$\sin\alpha=\frac{12}{13},\qquad \sin\beta=\frac{3}{5}\;.$$

Now we need the cosines of the same angles. We state the Pythagorean identity first: for any angle $$\theta$$, $$\sin^{2}\theta+\cos^{2}\theta=1\;.$$ Using it for $$\alpha$$ we obtain

$$\cos\alpha=\sqrt{1-\sin^{2}\alpha}=\sqrt{1-\Bigl(\frac{12}{13}\Bigr)^{2}} =\sqrt{1-\frac{144}{169}} =\sqrt{\frac{25}{169}} =\frac{5}{13}\;.$$

Applying the same identity to $$\beta$$ gives

$$\cos\beta=\sqrt{1-\sin^{2}\beta}=\sqrt{1-\Bigl(\frac{3}{5}\Bigr)^{2}} =\sqrt{1-\frac{9}{25}} =\sqrt{\frac{16}{25}} =\frac{4}{5}\;.$$

Our required expression is $$\alpha-\beta\;.$$ To find it we use the sine of the difference of two angles. We first recall the formula:

$$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta\;.$$

Substituting the values already obtained, we have

$$\sin(\alpha-\beta)=\Bigl(\frac{12}{13}\Bigr)\Bigl(\frac{4}{5}\Bigr) -\Bigl(\frac{5}{13}\Bigr)\Bigl(\frac{3}{5}\Bigr) =\frac{48}{65}-\frac{15}{65} =\frac{33}{65}\;.$$

Because $$\alpha-\beta$$ is the difference of two first-quadrant angles with $$\alpha \gt \beta,$$ the result is still an acute angle. Hence the principal value of $$\sin^{-1}$$ applies and we may write

$$\alpha-\beta=\sin^{-1}\frac{33}{65}\;.$$

So far we have $$\sin^{-1}\frac{12}{13}-\sin^{-1}\frac{3}{5} =\sin^{-1}\frac{33}{65}\;.$$

This exact form does not appear among the options, therefore we transform it. We make use of the co-function relation that links sine and cosine:

For any angle $$\theta$$ in $$[-\tfrac{\pi}{2},\tfrac{\pi}{2}],$$ $$\sin\theta=\cos\Bigl(\frac{\pi}{2}-\theta\Bigr)\;,$$ and, taking inverses, $$\sin^{-1}x=\frac{\pi}{2}-\cos^{-1}x\quad\text{for }-1\le x\le1\;.$$

Instead of rewriting $$\sin^{-1}\frac{33}{65}$$ in terms of $$\cos^{-1},$$ we notice there is an option of the form $$\tfrac{\pi}{2}-\sin^{-1}(\text{something})$$. Hence we try to express $$\sin^{-1}\frac{33}{65}$$ as $$\tfrac{\pi}{2}-\sin^{-1}\!y$$ for a suitable $$y$$.

Set $$\theta=\sin^{-1}\frac{33}{65}\;.$$ Then $$\sin\theta=\frac{33}{65}$$ and $$0\lt\theta\lt\frac{\pi}{2}\;.$$ Using the Pythagorean identity once again, we compute

$$\cos\theta=\sqrt{1-\sin^{2}\theta} =\sqrt{1-\Bigl(\frac{33}{65}\Bigr)^{2}} =\sqrt{1-\frac{1089}{4225}} =\sqrt{\frac{3136}{4225}} =\frac{56}{65}\;.$$

But $$\cos\theta=\sin\!\Bigl(\frac{\pi}{2}-\theta\Bigr)\;,$$ so

$$\frac{\pi}{2}-\theta=\sin^{-1}\frac{56}{65}\;.$$

Solving for $$\theta,$$ we get

$$\theta=\frac{\pi}{2}-\sin^{-1}\frac{56}{65}\;.$$

Remembering that $$\theta=\alpha-\beta,$$ we have proved

$$\sin^{-1}\frac{12}{13}-\sin^{-1}\frac{3}{5} =\frac{\pi}{2}-\sin^{-1}\frac{56}{65}\;.$$

This expression is exactly the one listed in Option 4.

Hence, the correct answer is Option 4.

We are given $$\alpha=\cos^{-1}\frac35$$ and $$\beta=\tan^{-1}\frac13$$ with $$0<\alpha,\beta<\dfrac{\pi}{2}$$, so both angles lie in the first quadrant. Because they are first-quadrant angles, every trigonometric ratio of each angle is positive.

From $$\alpha=\cos^{-1}\dfrac35$$ we have, by definition of the inverse cosine,

$$\cos\alpha=\dfrac35.$$

To find $$\sin\alpha$$ and $$\tan\alpha$$ we use the Pythagorean identity

$$\sin^2\theta+\cos^2\theta=1.$$

Substituting $$\cos\alpha=\dfrac35$$ gives

$$\sin^2\alpha+ \left(\dfrac35\right)^2 = 1 \quad\Longrightarrow\quad \sin^2\alpha + \dfrac9{25}=1.$$

Moving the fraction to the right side, we get

$$\sin^2\alpha = 1-\dfrac9{25}=\dfrac{25}{25}-\dfrac9{25}=\dfrac{16}{25}.$$

Taking the positive square root (because $$\alpha$$ is in the first quadrant) gives

$$\sin\alpha=\dfrac45.$$

Now $$\tan\alpha$$ can be written as

$$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\dfrac45}{\dfrac35}=\dfrac43.$$

Next consider $$\beta=\tan^{-1}\dfrac13.$$ By definition,

$$\tan\beta=\dfrac13.$$

To compute $$\sin\beta$$ and $$\cos\beta$$, imagine a right-angled triangle in which the side opposite $$\beta$$ is $$1$$ and the side adjacent to $$\beta$$ is $$3$$. The hypotenuse is obtained from the Pythagorean theorem:

$$\text{hypotenuse}=\sqrt{1^{2}+3^{2}}=\sqrt{1+9}=\sqrt{10}.$$

Therefore

$$\sin\beta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac1{\sqrt{10}}, \qquad \cos\beta=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac3{\sqrt{10}}.$$

We now need $$\alpha-\beta.$$ To relate this difference to a single trigonometric ratio we use the sine difference formula:

$$\sin(A-B)=\sin A\cos B-\cos A\sin B.$$

Taking $$A=\alpha$$ and $$B=\beta$$ gives

$$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta.$$

We substitute the values already found:

$$\sin(\alpha-\beta)=\left(\dfrac45\right)\!\left(\dfrac3{\sqrt{10}}\right)-\left(\dfrac35\right)\!\left(\dfrac1{\sqrt{10}}\right).$$

Multiplying numerators and denominators separately, we obtain

$$\sin(\alpha-\beta)=\dfrac{4\cdot3}{5\sqrt{10}}-\dfrac{3\cdot1}{5\sqrt{10}} =\dfrac{12}{5\sqrt{10}}-\dfrac{3}{5\sqrt{10}} =\dfrac{12-3}{5\sqrt{10}} =\dfrac9{5\sqrt{10}}.$$

So

$$\sin(\alpha-\beta)=\dfrac9{5\sqrt{10}}.$$

Because $$0<\alpha<\dfrac{\pi}{2}$$ and $$0<\beta<\dfrac{\pi}{2},$$ their difference also satisfies $$0<\alpha-\beta<\dfrac{\pi}{2}.$$ In this interval the arcsine function is single-valued, hence

$$\alpha-\beta=\sin^{-1}\!\left(\dfrac9{5\sqrt{10}}\right).$$

Looking at the options, this matches exactly Option C.

Hence, the correct answer is Option C.

We have to evaluate the quantity

$$\cot\!\left(\;\sum_{n=1}^{19} \cot^{-1}\!\left(1+\sum_{p=1}^{n}2p\right)\right).$$

First let us simplify the inner summation. The sum of the first $$n$$ even natural numbers can be written as $$2$$ times the sum of the first $$n$$ natural numbers. Using the well-known formula

$$1+2+\dots +n=\frac{n(n+1)}{2},$$

we get

$$\sum_{p=1}^{n}2p \;=\;2\left(\frac{n(n+1)}{2}\right)=n(n+1).$$

So the expression inside each $$\cot^{-1}$$ becomes

$$1+\sum_{p=1}^{n}2p = 1+n(n+1)=n^2+n+1.$$

Therefore the entire angle whose cotangent is being taken is

$$S=\sum_{n=1}^{19}\cot^{-1}\!\bigl(n^2+n+1\bigr).$$

To telescope this sum, it is easier to convert $$\cot^{-1}$$ to $$\tan^{-1}$$. Recall that for any positive $$x$$ we have

$$\cot^{-1}x = \tan^{-1}\!\left(\frac1x\right).$$

Applying this to each term,

$$\cot^{-1}\!\bigl(n^2+n+1\bigr)=\tan^{-1}\!\left(\frac1{n^2+n+1}\right).$$

Now we invoke the subtraction formula for $$\tan^{-1}$$. For real numbers $$\alpha$$ and $$\beta$$,

$$\tan^{-1}\alpha-\tan^{-1}\beta=\tan^{-1}\!\left(\frac{\alpha-\beta}{1+\alpha\beta}\right).$$

If we choose $$\alpha=n+1$$ and $$\beta=n$$, then

$$\tan^{-1}(n+1)-\tan^{-1}n=\tan^{-1}\!\left(\frac{(n+1)-n}{1+n(n+1)}\right) =\tan^{-1}\!\left(\frac1{n^2+n+1}\right).$$

This is exactly the form we obtained above. Hence, for every $$n\ge1$$,

$$\cot^{-1}\!\bigl(n^2+n+1\bigr)=\tan^{-1}(n+1)-\tan^{-1}n.$$

Substituting this identity into the sum for $$S$$ gives

$$S=\sum_{n=1}^{19}\Bigl[\tan^{-1}(n+1)-\tan^{-1}n\Bigr].$$

Observe that this is a telescoping series: consecutive terms cancel pairwise.

Explicitly, writing out the first few terms,

$$\bigl[\tan^{-1}2-\tan^{-1}1\bigr]+\bigl[\tan^{-1}3-\tan^{-1}2\bigr] +\dots+\bigl[\tan^{-1}20-\tan^{-1}19\bigr].$$

All the intermediate $$\tan^{-1}2,\tan^{-1}3,\dots,\tan^{-1}19$$ terms cancel, leaving only

$$S=\tan^{-1}20-\tan^{-1}1.$$

Next we need $$\cot S$$. First find $$\tan S$$ using the subtraction formula for tangent:

For angles $$A$$ and $$B$$,

$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\,\tan B}.$$

Taking $$A=\tan^{-1}20$$ (so $$\tan A=20$$) and $$B=\tan^{-1}1$$ (so $$\tan B=1$$), we have

$$\tan S=\tan\!\bigl(\tan^{-1}20-\tan^{-1}1\bigr) =\frac{20-1}{1+20\cdot1} =\frac{19}{21}.$$

Finally, since $$\cot S$$ is the reciprocal of $$\tan S$$,

$$\cot S=\frac{1}{\tan S}=\frac{21}{19}.$$

Hence, the correct answer is Option A.

We are given the equation

$$\cos^{-1}\!\left(\frac{2}{3x}\right)\;+\;\cos^{-1}\!\left(\frac{3}{4x}\right)\;=\;\frac{\pi}{2},\qquad x\;>\;\frac{3}{4}.$$

To remove the inverse-cosine symbols, we recall a standard trigonometric fact:

Formula stated first. If $$\cos^{-1}u+\cos^{-1}v=\frac{\pi}{2}$$ with $$0\le u\le1$$ and $$0\le v\le1,$$ then

$$u^{2}+v^{2}=1.$$

We justify this briefly. Write $$A=\cos^{-1}u$$ and $$B=\cos^{-1}v,$$ so that $$A+B=\frac{\pi}{2}.$$ Taking cosine on both sides,

$$\cos(A+B)=\cos\!\left(\frac{\pi}{2}\right)=0.$$

Using the cosine addition formula $$\cos(A+B)=\cos A\cos B-\sin A\sin B,$$ we have

$$u\;v-\sqrt{1-u^{2}}\;\sqrt{1-v^{2}}=0.$$

Thus

$$u\,v=\sqrt{1-u^{2}}\;\sqrt{1-v^{2}}.$$

Squaring both sides yields

$$u^{2}v^{2}=(1-u^{2})(1-v^{2}) =1-u^{2}-v^{2}+u^{2}v^{2},$$

and cancelling $$u^{2}v^{2}$$ from both sides gives the promised relation

$$u^{2}+v^{2}=1.$$

Now we identify

$$u=\frac{2}{3x},\qquad v=\frac{3}{4x}.$$

We substitute these into $$u^{2}+v^{2}=1.$$

First compute $$u^{2}$$ and $$v^{2}:$$

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

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

Adding them,

$$u^{2}+v^{2} =\frac{4}{9x^{2}}+\frac{9}{16x^{2}} =\frac{4\cdot16+9\cdot9}{144\,x^{2}} =\frac{64+81}{144\,x^{2}} =\frac{145}{144\,x^{2}}.$$

Setting this equal to $$1$$ (by the identity just derived) we obtain

$$\frac{145}{144\,x^{2}}=1.$$

Now isolate $$x^{2}:$$

$$x^{2}=\frac{145}{144}.$$

Taking the positive square root (because $$x>\frac{3}{4}>0$$),

$$x=\frac{\sqrt{145}}{12}.$$

We compare with the listed choices and find that this value matches Option D.

Hence, the correct answer is Option D.

We are given the relation $$\cos^{-1}x-\cos^{-1}\frac{y}{2}=\alpha$$.

Let us introduce two auxiliary angles to translate the inverse-cosine statement into ordinary cosine values. Put

$$\theta=\cos^{-1}x \qquad\text{and}\qquad \phi=\cos^{-1}\frac{y}{2}.$$

By definition of the inverse cosine we immediately have

$$x=\cos\theta,\qquad \frac{y}{2}=\cos\phi\;\;\Longrightarrow\;\;y=2\cos\phi.$$

The given equation now reads

$$\theta-\phi=\alpha\;\;\Longrightarrow\;\;\theta=\phi+\alpha.$$

Our target expression is

$$4x^{2}-4xy\cos\alpha+y^{2}.$$

Substituting $$x=\cos\theta$$ and $$y=2\cos\phi$$ we get

$$4(\cos\theta)^{2}-4(\cos\theta)(2\cos\phi)\cos\alpha+(2\cos\phi)^{2}.$$

Simplifying the numerical coefficients,

$$=4\cos^{2}\theta-8\cos\theta\cos\phi\cos\alpha+4\cos^{2}\phi.$$

Every term now carries a common factor 4, so we factor it out:

$$=4\Bigl(\cos^{2}\theta-2\cos\theta\cos\phi\cos\alpha+\cos^{2}\phi\Bigr).$$

We must evaluate the bracket. To do so we replace $$\theta$$ by $$\phi+\alpha$$ because $$\theta=\phi+\alpha.$$ First recall the compound-angle formula

$$\cos(A+B)=\cos A\cos B-\sin A\sin B.$$

Applying it to $$\cos\theta=\cos(\phi+\alpha)$$ yields

$$\cos\theta=\cos(\phi+\alpha)=\cos\phi\cos\alpha-\sin\phi\sin\alpha.$$

For brevity let us denote

$$C=\cos\theta,\qquad D=\cos\phi,\qquad S=\sin\phi.$$

Then $$C=D\cos\alpha-S\sin\alpha$$ and the bracket becomes

$$C^{2}-2CD\cos\alpha+D^{2}.$$

Compute $$C^{2}$$ first:

$$C^{2}=(D\cos\alpha-S\sin\alpha)^{2}=D^{2}\cos^{2}\alpha-2DS\cos\alpha\sin\alpha+S^{2}\sin^{2}\alpha.$$

Now insert this into the bracket:

$$$ \begin{aligned} &\;\;C^{2}-2CD\cos\alpha+D^{2}\\ &= \bigl(D^{2}\cos^{2}\alpha-2DS\cos\alpha\sin\alpha+S^{2}\sin^{2}\alpha\bigr) -2\bigl(D\cos\alpha-S\sin\alpha\bigr)D\cos\alpha +D^{2}. \end{aligned} $$$

Work out the middle product:

$$$ -2CD\cos\alpha=-2\bigl(D\cos\alpha-S\sin\alpha\bigr)D\cos\alpha =-2D^{2}\cos^{2}\alpha+2DS\cos\alpha\sin\alpha. $$$

Adding everything term by term:

$$$ \begin{aligned} C^{2}-2CD\cos\alpha+D^{2}&= \underbrace{D^{2}\cos^{2}\alpha}_{(1)} -\underbrace{2DS\cos\alpha\sin\alpha}_{(2)} +\underbrace{S^{2}\sin^{2}\alpha}_{(3)} \\ &\quad -\underbrace{2D^{2}\cos^{2}\alpha}_{(4)} +\underbrace{2DS\cos\alpha\sin\alpha}_{(5)} +\underbrace{D^{2}}_{(6)}. \end{aligned} $$$

The cross terms (2) and (5) cancel each other, leaving

$$-D^{2}\cos^{2}\alpha+S^{2}\sin^{2}\alpha+D^{2}.$$\p>

Factor out $$\sin^{2}\alpha$$ from the two compatible terms and remember $$\sin^{2}\alpha=1-\cos^{2}\alpha$$:

$$$ \begin{aligned} &=D^{2}\bigl(1-\cos^{2}\alpha\bigr)+S^{2}\sin^{2}\alpha\\ &=\sin^{2}\alpha\bigl(D^{2}+S^{2}\bigr). \end{aligned} $$$

But $$D^{2}+S^{2}=\cos^{2}\phi+\sin^{2}\phi=1,$$ so the entire bracket collapses beautifully to

$$\sin^{2}\alpha.$$

Returning to the full expression we earlier factored,

$$4\Bigl(\cos^{2}\theta-2\cos\theta\cos\phi\cos\alpha+\cos^{2}\phi\Bigr)=4\sin^{2}\alpha.$$

This value is completely independent of $$x$$ and $$y$$, depending only on the given constant angle $$\alpha$$.

Hence, the correct answer is Option D.

We recall the principal value ranges of the inverse trigonometric functions:

$$\sin^{-1}(t)\in\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right] \qquad\text{and}\qquad \cos^{-1}(t)\in\left[0,\,\pi\right].$$

Let $$\theta=10\;{\rm radians}.$$ Because $$2\pi\approx6.283$$ and $$4\pi\approx12.566,$$ the value $$\theta=10$$ satisfies

$$3\pi\;(\approx9.425)<10<4\pi\;(\approx12.566).$$

Evaluating $$x=\sin^{-1}(\sin10).$$

The sine function has period $$2\pi,$$ so $$\sin\theta=\sin(\theta-2n\pi)$$ for any integer $$n.$$ Choose $$n=1:$$

$$\theta-2\pi=10-2\pi\approx3.717.$$

This result still lies outside the principal interval $$\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right],$$ so subtract one more period:

$$\theta-4\pi=10-4\pi\approx-2.566.$$

The number $$-2.566$$ lies between $$-\pi$$ and $$-\dfrac{\pi}{2},$$ not yet in the principal range. Using $$\sin(-\alpha)=-\sin\alpha$$ and the symmetry $$\sin(\pi-\alpha)=\sin\alpha,$$ the standard result for this situation is

$$\sin^{-1}(\sin\theta)=(2n+1)\pi-\theta$$ when $$\theta\in\left[2n\pi+\dfrac{\pi}{2},\,2n\pi+\dfrac{3\pi}{2}\right].$$

Because $$10\in\left[2\pi+\dfrac{\pi}{2},\,2\pi+\dfrac{3\pi}{2}\right]$$ with $$n=1,$$ we obtain

$$x=(2\!\cdot\!1+1)\pi-10=3\pi-10.$$

Numerically, $$3\pi-10\approx-0.575,$$ which indeed lies inside $$\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].$$

Evaluating $$y=\cos^{-1}(\cos10).$$

The cosine function also has period $$2\pi,$$ and the standard formula is

$$ \cos^{-1}(\cos\theta)= \begin{cases} \theta-2m\pi, & \theta\in[2m\pi,\,2m\pi+\pi],\\[6pt] 2(m+1)\pi-\theta, & \theta\in[2m\pi+\pi,\,2m\pi+2\pi], \end{cases} $$ where $$m$$ is an integer.

Because $$10\in[3\pi,\,4\pi]=[2\!\cdot\!1\pi+\pi,\,2\!\cdot\!1\pi+2\pi]$$ with $$m=1,$$ we are in the second case. Substituting $$m=1$$ gives

$$y=2(m+1)\pi-\theta=2\!\cdot\!2\pi-10=4\pi-10.$$

Numerically, $$4\pi-10\approx2.566,$$ which lies in the required interval $$[0,\pi].$$

Finding $$y-x.$$ We have

$$ y-x=(4\pi-10)-(3\pi-10)=4\pi-10-3\pi+10=\pi. $$

Thus $$y-x=\pi.$$ Hence, the correct answer is Option B.

Let $$\cot^{-1}(1 + x) = \theta$$. This means $$\cot \theta = \frac{1 + x}{1}$$.

In a right triangle with adjacent side 1+x and opposite side 1,

Hypotenuse $$= \sqrt{1^2 + (1 + x)^2} = \sqrt{1 + 1 + x^2 + 2x} = \sqrt{x^2 + 2x + 2}$$

Therefore, $$\sin \theta = \frac{1}{\sqrt{x^2 + 2x + 2}}$$

Let $$\tan^{-1} x = \phi$$. This means $$\tan \phi = \frac{x}{1}$$.

In a right triangle where the opposite side is x and the adjacent side is 1,

Hypotenuse $$= \sqrt{1^2 + x^2} = \sqrt{1 + x^2}$$

Therefore, $$\cos \phi = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{1}{\sqrt{1 + x^2}}$$

Now substitute the simplified forms back into the original equation:

$$\frac{1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{1 + x^2}}$$

$$\frac{1}{x^2 + 2x + 2} = \frac{1}{1 + x^2}$$

$$1 + x^2 = x^2 + 2x + 2$$

$$1 = 2x + 2$$

$$\therefore\ $$ $$x = -\frac{1}{2}$$

We have to evaluate the expression

$$\tan^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]$$

where the domain condition $$|x|\lt \dfrac12,\;x\neq0$$ is given so that every square-root and inverse-trig quantity involved is real and single-valued in its principal branch.

First we recall the tangent addition formula written in the form

$$\tan\!\left(\frac{\pi}{4}+y\right)=\frac{1+\tan y}{1-\tan y}.$$

Comparing this standard result with the fraction that appears inside the inverse tangent, we try to bring that fraction to the pattern $$\dfrac{1+\tan y}{1-\tan y}.$$

Let us denote

$$A=\sqrt{1+x^2},\qquad B=\sqrt{1-x^2}.$$

Then the argument of $$\tan^{-1}$$ is

$$\frac{A+B}{A-B}.$$

We divide the numerator and the denominator by $$A$$ (which is positive, because $$1+x^2\gt 0$$):

$$\frac{A+B}{A-B}=\frac{\dfrac{A}{A}+\dfrac{B}{A}}{\dfrac{A}{A}-\dfrac{B}{A}}=\frac{1+\dfrac{B}{A}}{1-\dfrac{B}{A}}.$$

Now set

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

Because $$0\lt |x|\lt \dfrac12$$, we have $$0\lt \frac{B}{A}\lt 1$$, so $$y$$ lies in the interval $$\left(0,\dfrac{\pi}{4}\right),$$ well inside the principal branch. With this choice we obtain

$$\frac{A+B}{A-B}=\frac{1+\tan y}{1-\tan y}=\tan\!\left(\frac{\pi}{4}+y\right).$$

Therefore

$$\tan^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]=\frac{\pi}{4}+y.$$

Our task is now reduced to expressing $$y$$ in terms of $$x$$. We already know

$$\tan y=\sqrt{\frac{1-x^2}{1+x^2}}.$$

To convert this to a cosine-inverse expression we use the double-angle identity

$$\cos 2y=\frac{1-\tan^2 y}{1+\tan^2 y}.$$

Compute the square of the tangent:

$$\tan^2 y=\frac{1-x^2}{1+x^2}.$$

Now substitute in the identity:

$$$ \begin{aligned} \cos 2y &=\frac{1-\dfrac{1-x^2}{1+x^2}}{1+\dfrac{1-x^2}{1+x^2}} =\frac{\dfrac{(1+x^2)-(1-x^2)}{1+x^2}}{\dfrac{(1+x^2)+(1-x^2)}{1+x^2}} =\frac{2x^2}{2} =x^2. \end{aligned} $$$

We have obtained $$\cos 2y=x^2.$$ The principal value of $$\cos^{-1}$$ lies in $$[0,\pi],$$ and because $$y\in\!\left(0,\dfrac{\pi}{4}\right)$$ we get $$2y\in\!\left(0,\dfrac{\pi}{2}\right)\subset[0,\pi],$$ so no ambiguity arises. Hence

$$2y=\cos^{-1}(x^2)\quad\Longrightarrow\quad y=\frac12\,\cos^{-1}(x^2).$$

Substituting this back into the earlier result gives

$$\tan^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]=\frac{\pi}{4}+\frac12\,\cos^{-1}(x^2).$$

Thus the given expression equals $$\dfrac{\pi}{4}+\dfrac{1}{2}\cos^{-1}x^2,$$ which exactly matches Option A.

Hence, the correct answer is Option A.

We are given the relation $$\tan^{-1}y=\tan^{-1}x+\tan^{-1}\!\left(\dfrac{2x}{1-x^{2}}\right)$$ with the condition $$|x|\lt \dfrac{1}{\sqrt{3}}$$ and we have to find an explicit expression for $$y$$.

First, recall the standard sum-of-inverse-tangents formula:

$$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\!\left(\dfrac{A+B}{1-AB}\right),$$

which is valid when the product $$AB\lt 1$$, a condition that will indeed be satisfied here because of the given bound on $$x$$.

We now apply the formula to the two inverse-tangent terms on the right-hand side. Identifying

$$A=x,\qquad B=\dfrac{2x}{1-x^{2}},$$

we can write

$$\tan^{-1}y=\tan^{-1}\!\left(\dfrac{A+B}{\;1-AB\;}\right).$$

So we need to compute both the numerator $$A+B$$ and the denominator $$1-AB$$ in explicit algebraic form.

Numerator:

$$$ A+B \;=\; x \;+\; \dfrac{2x}{1-x^{2}} = x\!\left(1\right) + x\!\left(\dfrac{2}{1-x^{2}}\right) = x\!\left(\dfrac{1-x^{2}}{1-x^{2}}\right) + x\!\left(\dfrac{2}{1-x^{2}}\right) = \dfrac{x\!\left(1-x^{2} + 2\right)}{1-x^{2}} = \dfrac{x\!\left(3 - x^{2}\right)}{1-x^{2}}. $$$

Denominator:

$$$ 1-AB = 1 - x\!\left(\dfrac{2x}{1-x^{2}}\right) = 1 - \dfrac{2x^{2}}{1-x^{2}} = \dfrac{1-x^{2}}{1-x^{2}} - \dfrac{2x^{2}}{1-x^{2}} = \dfrac{1 - x^{2} - 2x^{2}}{1-x^{2}} = \dfrac{1 - 3x^{2}}{1-x^{2}}. $$$

Putting numerator and denominator together:

$$$ \dfrac{A+B}{1-AB} = \dfrac{\dfrac{x(3 - x^{2})}{1-x^{2}}}{\dfrac{1 - 3x^{2}}{1-x^{2}}} = \dfrac{x(3 - x^{2})}{1 - 3x^{2}}, $$$

because the common factor $$1-x^{2}$$ cancels out completely.

Hence we have obtained

$$\tan^{-1}y=\tan^{-1}\!\left(\dfrac{x(3 - x^{2})}{1 - 3x^{2}}\right).$$

Since the inverse-tangent function is one-to-one in its principal branch, the arguments must be equal, and therefore

$$y=\dfrac{x(3 - x^{2})}{1 - 3x^{2}} =\dfrac{3x - x^{3}}{1 - 3x^{2}}.$$

This matches Option B in the list provided.

Hence, the correct answer is Option B.

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

If $$x > 1$$: $$\sin^{-1} \left( \frac{2x}{1+x^2} \right) = \pi - 2 \tan^{-1} x$$

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

$$f(x) = 2 \tan^{-1} x + \left( \pi - 2 \tan^{-1} x \right)$$ 

$$f(x) = \pi$$

$$f(5) = \pi$$

We need to find the principal value of $$\tan^{-1}\left(\cot\frac{43\pi}{4}\right)$$.

First, recall that $$\cot\theta = \tan\left(\frac{\pi}{2} - \theta\right)$$. So, we can write:

$$\cot\frac{43\pi}{4} = \tan\left(\frac{\pi}{2} - \frac{43\pi}{4}\right)$$

Now, compute the expression inside the tangent:

$$\frac{\pi}{2} - \frac{43\pi}{4} = \frac{2\pi}{4} - \frac{43\pi}{4} = \frac{2\pi - 43\pi}{4} = \frac{-41\pi}{4}$$

So, $$\cot\frac{43\pi}{4} = \tan\left(-\frac{41\pi}{4}\right)$$.

Since tangent is an odd function, $$\tan(-\theta) = -\tan\theta$$, so:

$$\tan\left(-\frac{41\pi}{4}\right) = -\tan\left(\frac{41\pi}{4}\right)$$

Thus, $$\cot\frac{43\pi}{4} = -\tan\left(\frac{41\pi}{4}\right)$$.

Next, simplify $$\tan\left(\frac{41\pi}{4}\right)$$. The tangent function has a period of $$\pi$$, so we reduce the angle modulo $$\pi$$.

Write $$\frac{41\pi}{4}$$ as:

$$\frac{41\pi}{4} = \frac{41}{4}\pi = 10\pi + \frac{\pi}{4}$$

because $$10\pi = \frac{40\pi}{4}$$, and $$\frac{41\pi}{4} - \frac{40\pi}{4} = \frac{\pi}{4}$$.

Since $$\tan(\theta + k\pi) = \tan\theta$$ for any integer $$k$$, and $$10\pi$$ is a multiple of $$\pi$$ (with $$k=10$$):

$$\tan\left(10\pi + \frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1$$

Therefore, $$\tan\left(\frac{41\pi}{4}\right) = 1$$, and:

$$\cot\frac{43\pi}{4} = -\tan\left(\frac{41\pi}{4}\right) = -1$$

So, the expression becomes:

$$\tan^{-1}\left(\cot\frac{43\pi}{4}\right) = \tan^{-1}(-1)$$

The principal value of $$\tan^{-1}(x)$$ lies in the interval $$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$. We know that $$\tan\left(-\frac{\pi}{4}\right) = -1$$, and $$-\frac{\pi}{4}$$ is within this interval. Hence, the principal value is $$-\frac{\pi}{4}$$.

Now, comparing with the options:

A. $$\frac{\pi}{4}$$

B. $$-\frac{\pi}{4}$$

C. $$\frac{3\pi}{4}$$

D. $$-\frac{3\pi}{4}$$

Hence, the correct answer is Option B.

Let $$f(x) = (\sin^{-1}x)^3 + (\cos^{-1}x)^3$$. Using the identity $$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$ for $$x \in [-1, 1]$$, let $$\sin^{-1}x = t$$, where $$t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$$.

$$f(x) = t^3 + \left(\frac{\pi}{2} - t\right)^3$$

$$A^3 + B^3 = (A+B)(A^2 - AB + B^2)$$: 

$$f(x) = \left(t + \frac{\pi}{2} - t\right) \left[t^2 - t\left(\frac{\pi}{2} - t\right) + \left(\frac{\pi}{2} - t\right)^2\right]$$

$$f(x) = \frac{\pi}{2} \left[t^2 - \frac{\pi t}{2} + t^2 + \frac{\pi^2}{4} - \pi t + t^2\right]$$

$$f(x) = \frac{\pi}{2} \left[3t^2 - \frac{3\pi t}{2} + \frac{\pi^2}{4}\right]$$

$$f(x) = \frac{3\pi}{2} \left[t^2 - \frac{\pi t}{2} + \frac{\pi^2}{12}\right]$$

$$f(x) = \frac{3\pi}{2} \left[\left(t - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{48}\right]$$

For $$x \in [-1, 1]$$, the range of $$t = \sin^{-1}x$$ is $$[-\frac{\pi}{2}, \frac{\pi}{2}]$$.

The range of $$\left(t - \frac{\pi}{4}\right)$$ is $$[-\frac{3\pi}{4}, \frac{\pi}{4}]$$.

Squaring this interval: $$0 \le \left(\sin^{-1}x - \frac{\pi}{4}\right)^2 \le \left(-\frac{3\pi}{4}\right)^2 = \frac{9\pi^2}{16}$$

Thus, Statement II is true.

Using the range of $$\left(t - \frac{\pi}{4}\right)^2$$ in the expression for $$f(x)$$:

Minimum value: $$f(x)_{min} = \frac{3\pi}{2} \left[0 + \frac{\pi^2}{48}\right] = \frac{\pi^3}{32}$$

Maximum value: $$f(x)_{max} = \frac{3\pi}{2} \left[\frac{9\pi^2}{16} + \frac{\pi^2}{48}\right] = \frac{3\pi}{2} \left[\frac{28\pi^2}{48}\right] = \frac{7\pi^3}{8}$$

The given equation $$(\sin^{-1}x)^3 + (\cos^{-1}x)^3 = a\pi^3$$ has a solution if $$a\pi^3 \in [\frac{\pi^3}{32}, \frac{7\pi^3}{8}]$$.

This implies $$\frac{1}{32} \le a \le \frac{7}{8}$$

In the context of the given statements, the condition $$a \ge \frac{1}{32}$$ identifies the lower bound necessary for a solution to exist. Hence, Statement I is false.

We are told that the three positive numbers $$x$$, $$y$$ and $$z$$ are in an arithmetic progression. By definition of an A.P. we therefore have

$$y=\dfrac{x+z}{2}.$$

Next, the three angles $$\tan^{-1}x$$, $$\tan^{-1}y$$ and $$\tan^{-1}z$$ are also in an arithmetic progression. For any three angles $$A,B,C$$ to be in A.P. we must have $$2B=A+C$$. Applying this to the present case gives

$$2\,\tan^{-1}y=\tan^{-1}x+\tan^{-1}z.$$

Now we take the tangent of both sides. Before doing so, let us recall the two standard trigonometric identities we shall need:

1. Sum formula for tangent: $$\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\,\tan\beta}.$$

2. Double-angle formula for tangent: $$\tan(2\theta)=\dfrac{2\tan\theta}{1-\tan^{2}\theta}.$$

Using the double-angle formula on the left-hand side with $$\theta=\tan^{-1}y$$ gives

$$\tan\!\bigl(2\,\tan^{-1}y\bigr)=\dfrac{2y}{1-y^{2}}.$$

Using the sum formula on the right-hand side with $$\alpha=\tan^{-1}x$$ and $$\beta=\tan^{-1}z$$ yields

$$\tan\!\bigl(\tan^{-1}x+\tan^{-1}z\bigr)=\dfrac{x+z}{1-xz}.$$

Equating the two tangent values obtained from the equality of the angles, we have

$$\dfrac{2y}{1-y^{2}}=\dfrac{x+z}{1-xz}.$$

But from the A.P. condition we already know $$x+z=2y,$$ so we substitute this into the numerator on the right-hand side:

$$\dfrac{2y}{1-y^{2}}=\dfrac{2y}{1-xz}.$$

Both numerators are equal and non-zero (all numbers are positive), so we can cancel the common factor $$2y$$, obtaining

$$\dfrac{1}{1-y^{2}}=\dfrac{1}{1-xz}.$$

Cross-multiplying gives

$$1-xz=1-y^{2}\quad\Longrightarrow\quad y^{2}=xz.$$

Thus $$y$$ is the geometric mean of $$x$$ and $$z$$, while we already have $$y$$ as the arithmetic mean of $$x$$ and $$z$$. Let us now combine the two conditions.

From the arithmetic mean relation $$y=\dfrac{x+z}{2}$$, squaring both sides gives

$$y^{2}=\dfrac{(x+z)^{2}}{4}.$$

Substituting the geometric-mean equality $$y^{2}=xz$$ into this squared expression, we get

$$xz=\dfrac{(x+z)^{2}}{4}.$$

Multiplying throughout by $$4$$ yields

$$4xz=(x+z)^{2}=x^{2}+2xz+z^{2}.$$

Bringing all terms to one side, we have

$$0=x^{2}+2xz+z^{2}-4xz=x^{2}-2xz+z^{2}=(x-z)^{2}.$$

The square of a real number is zero only when the number itself is zero. Hence

$$x-z=0\quad\Longrightarrow\quad x=z.$$

Substituting $$x=z$$ back into $$y=\dfrac{x+z}{2}$$ gives

$$y=\dfrac{x+x}{2}=x.$$

Therefore $$x=y=z.$$

Looking at the options, we see that Option C states exactly this equality.

Hence, the correct answer is Option C.

The given equation is $$\sin\left(\cot^{-1}(1+x)\right) = \cos\left(\tan^{-1}x\right)$$. To solve for $$x$$, express both sides in terms of $$x$$. Consider the left side: $$\sin\left(\cot^{-1}(1+x)\right)$$. Let $$\alpha = \cot^{-1}(1+x)$$, so $$\cot \alpha = 1+x$$. In a right triangle, $$\cot \alpha = \frac{\text{adjacent}}{\text{opposite}}$$, so set adjacent = $$1+x$$ and opposite = 1. The hypotenuse is $$\sqrt{(1+x)^2 + 1^2} = \sqrt{1 + 2x + x^2 + 1} = \sqrt{x^2 + 2x + 2}$$. Thus, $$\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 2x + 2}}$$. Now consider the right side: $$\cos\left(\tan^{-1}x\right)$$. Let $$\beta = \tan^{-1}x$$, so $$\tan \beta = x$$. In a right triangle, $$\tan \beta = \frac{\text{opposite}}{\text{adjacent}}$$, so set opposite = $$x$$ and adjacent = 1. The hypotenuse is $$\sqrt{x^2 + 1^2} = \sqrt{x^2 + 1}$$. Thus, $$\cos \beta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}}$$. The equation becomes: $$$ \frac{1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{x^2 + 1}} $$$ Since the denominators are positive, take reciprocals: $$$ \sqrt{x^2 + 2x + 2} = \sqrt{x^2 + 1} $$$ Square both sides to eliminate the square roots: $$$ (\sqrt{x^2 + 2x + 2})^2 = (\sqrt{x^2 + 1})^2 $$$ $$$ x^2 + 2x + 2 = x^2 + 1 $$$ Subtract $$x^2$$ from both sides: $$$ 2x + 2 = 1 $$$ Subtract 2 from both sides: $$$ 2x = -1 $$$ Divide by 2: $$$ x = -\frac{1}{2} $$$ Now, verify if $$x = -\frac{1}{2}$$ satisfies the original equation. Substitute $$x = -\frac{1}{2}$$: Left side: $$\sin\left(\cot^{-1}\left(1 + \left(-\frac{1}{2}\right)\right)\right) = \sin\left(\cot^{-1}\left(\frac{1}{2}\right)\right)$$. Let $$\theta = \cot^{-1}\left(\frac{1}{2}\right)$$, so $$\cot \theta = \frac{1}{2}$$. In a right triangle, adjacent = 1, opposite = 2, hypotenuse = $$\sqrt{1^2 + 2^2} = \sqrt{5}$$. Thus, $$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}}$$. Right side: $$\cos\left(\tan^{-1}\left(-\frac{1}{2}\right)\right)$$. Let $$\phi = \tan^{-1}\left(-\frac{1}{2}\right)$$, so $$\tan \phi = -\frac{1}{2}$$. Since cosine is positive in the fourth quadrant, consider a triangle with opposite = $$\frac{1}{2}$$ (magnitude), adjacent = 1, hypotenuse = $$\sqrt{\left(\frac{1}{2}\right)^2 + 1^2} = \sqrt{\frac{1}{4} + 1} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$$. Thus, $$\cos \phi = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\frac{\sqrt{5}}{2}} = \frac{2}{\sqrt{5}}$$. Both sides equal $$\frac{2}{\sqrt{5}}$$, so $$x = -\frac{1}{2}$$ satisfies the equation. Now, check the other options to confirm: - Option B: $$x = 1$$ Left side: $$\sin\left(\cot^{-1}(1+1)\right) = \sin\left(\cot^{-1}(2)\right)$$. Let $$\theta = \cot^{-1}(2)$$, so $$\cot \theta = 2$$. Adjacent = 2, opposite = 1, hypotenuse = $$\sqrt{2^2 + 1^2} = \sqrt{5}$$, so $$\sin \theta = \frac{1}{\sqrt{5}}$$. Right side: $$\cos\left(\tan^{-1}(1)\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$. $$\frac{1}{\sqrt{5}} \neq \frac{1}{\sqrt{2}}$$, so not equal. - Option C: $$x = 0$$ Left side: $$\sin\left(\cot^{-1}(1+0)\right) = \sin\left(\cot^{-1}(1)\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$. Right side: $$\cos\left(\tan^{-1}(0)\right) = \cos(0) = 1$$. $$\frac{\sqrt{2}}{2} \neq 1$$, so not equal. - Option D: $$x = \frac{1}{2}$$ Left side: $$\sin\left(\cot^{-1}\left(1 + \frac{1}{2}\right)\right) = \sin\left(\cot^{-1}\left(\frac{3}{2}\right)\right)$$. Let $$\theta = \cot^{-1}\left(\frac{3}{2}\right)$$, so $$\cot \theta = \frac{3}{2}$$. Adjacent = $$\frac{3}{2}$$, opposite = 1, hypotenuse = $$\sqrt{\left(\frac{3}{2}\right)^2 + 1^2} = \sqrt{\frac{9}{4} + 1} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$$, so $$\sin \theta = \frac{1}{\frac{\sqrt{13}}{2}} = \frac{2}{\sqrt{13}}$$. Right side: $$\cos\left(\tan^{-1}\left(\frac{1}{2}\right)\right)$$. Let $$\phi = \tan^{-1}\left(\frac{1}{2}\right)$$, so $$\tan \phi = \frac{1}{2}$$. Adjacent = 1, opposite = $$\frac{1}{2}$$, hypotenuse = $$\sqrt{1^2 + \left(\frac{1}{2}\right)^2} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$$, so $$\cos \phi = \frac{1}{\frac{\sqrt{5}}{2}} = \frac{2}{\sqrt{5}}$$. $$\frac{2}{\sqrt{13}} \neq \frac{2}{\sqrt{5}}$$, so not equal. Only $$x = -\frac{1}{2}$$ satisfies the equation, which corresponds to option A. Hence, the correct answer is Option A.

We are given that $$ x \in (0, 1) $$ and need to find the set of all $$ x $$ such that $$ \sin^{-1} x > \cos^{-1} x $$.

Recall the identity: for any $$ x \in [-1, 1] $$, $$ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} $$. Therefore, we can express $$ \cos^{-1} x $$ as $$ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x $$.

Substitute this into the inequality:

$$ \sin^{-1} x > \cos^{-1} x $$

$$ \sin^{-1} x > \frac{\pi}{2} - \sin^{-1} x $$

Add $$ \sin^{-1} x $$ to both sides:

$$ \sin^{-1} x + \sin^{-1} x > \frac{\pi}{2} $$

$$ 2 \sin^{-1} x > \frac{\pi}{2} $$

Divide both sides by 2:

$$ \sin^{-1} x > \frac{\pi}{4} $$

Since $$ x \in (0, 1) $$, $$ \sin^{-1} x $$ lies in $$ (0, \frac{\pi}{2}) $$, where the sine function is strictly increasing. Therefore, taking sine on both sides preserves the inequality:

$$ \sin(\sin^{-1} x) > \sin\left(\frac{\pi}{4}\right) $$

$$ x > \sin\left(\frac{\pi}{4}\right) $$

$$ x > \frac{1}{\sqrt{2}} $$

Also, since $$ x \in (0, 1) $$, we have $$ x < 1 $$. Thus, $$ x $$ must satisfy $$ \frac{1}{\sqrt{2}} < x < 1 $$, which is the interval $$ \left( \frac{1}{\sqrt{2}}, 1 \right) $$.

Now, verify the boundaries:

• At $$ x = \frac{1}{\sqrt{2}} $$, $$ \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} $$ and $$ \cos^{-1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} $$, so equality holds, not inequality.
• As $$ x $$ approaches 1 from the left, $$ \sin^{-1} x $$ approaches $$ \frac{\pi}{2} $$ and $$ \cos^{-1} x $$ approaches 0, so $$ \sin^{-1} x > \cos^{-1} x $$ holds.

Comparing with the options:

• A: $$ \left( \frac{1}{2}, \frac{1}{\sqrt{2}} \right) $$ → This includes values less than $$ \frac{1}{\sqrt{2}} $$, which do not satisfy the inequality.
• B: $$ \left( \frac{1}{\sqrt{2}}, 1 \right) $$ → Matches our solution.
• C: $$ (0, 1) $$ → Includes values less than or equal to $$ \frac{1}{\sqrt{2}} $$, which do not satisfy.
• D: $$ \left( 0, \frac{\sqrt{3}}{2} \right) $$ → $$ \frac{\sqrt{3}}{2} \approx 0.866 $$ and $$ \frac{1}{\sqrt{2}} \approx 0.707 $$, so this interval includes values below $$ \frac{1}{\sqrt{2}} $$ and excludes values above $$ \frac{\sqrt{3}}{2} $$ that are in $$ \left( \frac{1}{\sqrt{2}}, 1 \right) $$, hence incorrect.

Hence, the correct answer is Option B.

The given sum is:

$$S = \tan^{-1}\left(\frac{1}{n^2+n+1}\right) + \tan^{-1}\left(\frac{1}{n^2+3n+3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+(n+19)(n+20)}\right)$$

First, we identify the general term. The last term has the denominator $$1 + (n+19)(n+20) = 1 + n^2 + 39n + 380 = n^2 + 39n + 381$$. Observing the denominators:

• The first term has denominator $$n^2 + n + 1$$.
• The second term has denominator $$n^2 + 3n + 3$$.

We notice that for each term, the denominator is of the form $$n^2 + (2k+1)n + (k^2 + k + 1)$$ where $$k$$ ranges from 0 to 19. Let's verify:

• For $$k = 0$$: $$n^2 + (2 \cdot 0 + 1)n + (0^2 + 0 + 1) = n^2 + n + 1$$, which matches the first term.
• For $$k = 1$$: $$n^2 + (2 \cdot 1 + 1)n + (1^2 + 1 + 1) = n^2 + 3n + 3$$, which matches the second term.
• For $$k = 19$$: $$n^2 + (2 \cdot 19 + 1)n + (19^2 + 19 + 1) = n^2 + 39n + (361 + 19 + 1) = n^2 + 39n + 381$$, which matches the last term.

Thus, the sum has 20 terms, from $$k = 0$$ to $$k = 19$$:

$$S = \sum_{k=0}^{19} \tan^{-1}\left(\frac{1}{n^2 + (2k+1)n + (k^2 + k + 1)}\right)$$

We can rewrite the denominator as:

$$n^2 + (2k+1)n + (k^2 + k + 1) = (n^2 + 2kn + k^2) + (n + k) + 1 = (n + k)^2 + (n + k) + 1$$

So each term becomes:

$$\tan^{-1}\left(\frac{1}{(n + k)^2 + (n + k) + 1}\right)$$

Now, recall the identity:

$$\tan^{-1}\left(\frac{1}{m^2 + m + 1}\right) = \tan^{-1}(m + 1) - \tan^{-1}(m)$$

This is verified by setting $$\theta = \tan^{-1}(m + 1) - \tan^{-1}(m)$$, so:

$$\tan \theta = \frac{(m + 1) - m}{1 + (m + 1)m} = \frac{1}{1 + m(m + 1)} = \frac{1}{m^2 + m + 1}$$

Thus, $$\theta = \tan^{-1}\left(\frac{1}{m^2 + m + 1}\right)$$.

Applying this identity with $$m = n + k$$, each term simplifies to:

$$\tan^{-1}\left(\frac{1}{(n + k)^2 + (n + k) + 1}\right) = \tan^{-1}((n + k) + 1) - \tan^{-1}(n + k) = \tan^{-1}(n + k + 1) - \tan^{-1}(n + k)$$

Therefore, the sum $$S$$ becomes:

$$S = \sum_{k=0}^{19} \left[ \tan^{-1}(n + k + 1) - \tan^{-1}(n + k) \right]$$

This is a telescoping series. Writing out the terms:

• For $$k = 0$$: $$\tan^{-1}(n + 1) - \tan^{-1}(n)$$
• For $$k = 1$$: $$\tan^{-1}(n + 2) - \tan^{-1}(n + 1)$$
• For $$k = 2$$: $$\tan^{-1}(n + 3) - \tan^{-1}(n + 2)$$
• $$\vdots$$
• For $$k = 19$$: $$\tan^{-1}(n + 20) - \tan^{-1}(n + 19)$$

Summing these, intermediate terms cancel:

$$S = -\tan^{-1}(n) + \tan^{-1}(n + 20)$$

Because $$\tan^{-1}(n + 1)$$ cancels with $$-\tan^{-1}(n + 1)$$, $$\tan^{-1}(n + 2)$$ cancels with $$-\tan^{-1}(n + 2)$$, and so on, up to $$\tan^{-1}(n + 19)$$.

So:

$$S = \tan^{-1}(n + 20) - \tan^{-1}(n)$$

Now, we need to find $$\tan S = \tan\left( \tan^{-1}(n + 20) - \tan^{-1}(n) \right)$$. Using the tangent subtraction formula:

$$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$

Here, $$A = \tan^{-1}(n + 20)$$, so $$\tan A = n + 20$$, and $$B = \tan^{-1}(n)$$, so $$\tan B = n$$. Substituting:

$$\tan S = \frac{(n + 20) - n}{1 + (n + 20) \cdot n} = \frac{20}{1 + n(n + 20)}$$

Simplifying the denominator:

$$1 + n(n + 20) = 1 + n^2 + 20n = n^2 + 20n + 1$$

Thus:

$$\tan S = \frac{20}{n^2 + 20n + 1}$$

Comparing with the options:

• A. $$\frac{20}{401 + 20n}$$
• B. $$\frac{n}{n^2 + 20n + 1}$$
• C. $$\frac{20}{n^2 + 20n + 1}$$
• D. $$\frac{n}{401 + 20n}$$

Option C matches the expression we obtained.

Hence, the correct answer is Option C.

Let the side lengths opposite to angles $$A,B,C$$ be $$a,b,c$$ respectively. The condition given is

$$\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=k,$$

where $$k$$ is a positive constant (because all sides are positive). From this we obtain three linear equations:

$$b+c = 11k \quad -(1)$$
$$c+a = 12k \quad -(2)$$
$$a+b = 13k \quad -(3)$$

First, add equations $$(1)$$ and $$(2)$$:

$$b+c + c+a = 11k + 12k \; \Rightarrow \; a+b+2c = 23k \quad -(4)$$

Subtract equation $$(3)$$ from $$(4)$$ to isolate $$c$$:

$$(a+b+2c) - (a+b) = 23k - 13k \; \Rightarrow \; 2c = 10k \; \Rightarrow \; c = 5k.$$

Insert $$c = 5k$$ into equation $$(2)$$ to get $$a$$:

$$5k + a = 12k \; \Rightarrow \; a = 7k.$$

Insert $$a = 7k$$ into equation $$(3)$$ to get $$b$$:

$$7k + b = 13k \; \Rightarrow \; b = 6k.$$

Hence the side lengths are proportional to $$7:6:5$$, i.e.

$$a = 7k,\; b = 6k,\; c = 5k.$$

To find $$\cos A$$, use the Law of Cosines:

$$\cos A = \frac{b^{2}+c^{2}-a^{2}}{2bc}.$$

Substitute the obtained side lengths:

$$\cos A = \frac{(6k)^{2} + (5k)^{2} - (7k)^{2}}{2\,(6k)\,(5k)} = \frac{36k^{2} + 25k^{2} - 49k^{2}}{60k^{2}} = \frac{12k^{2}}{60k^{2}} = \frac{1}{5}.$$

Therefore, $$\cos A = \dfrac{1}{5}.$$

For three coplanar forces to be in equilibrium, their vectors must form a closed triangle.
Equivalently, if the magnitudes are $$P,Q,R$$ and the angles between the directions of $$\vec Q,\vec R;\; \vec R,\vec P;\; \vec P,\vec Q$$ are $$\theta_A,\theta_B,\theta_C$$ respectively, then by the sine rule for the triangle of forces

$$\frac{P}{\sin\theta_A}= \frac{Q}{\sin\theta_B}= \frac{R}{\sin\theta_C}\;.\qquad -(1)$$

In this problem the forces act along the internal angle-bisectors $$IA,IB,IC$$ of $$\triangle ABC$$. Hence:

• the angle between $$IB$$ and $$IC$$ at the incentre $$I$$ equals $$\theta_A=90^\circ+\frac{A}{2}$$,
• the angle between $$IC$$ and $$IA$$ equals $$\theta_B=90^\circ+\frac{B}{2}$$,
• the angle between $$IA$$ and $$IB$$ equals $$\theta_C=90^\circ+\frac{C}{2}$$.

Using $$\sin(90^\circ+\phi)=\cos\phi$$, equation $$(1)$$ becomes

$$\frac{P}{\cos\frac{A}{2}}=\frac{Q}{\cos\frac{B}{2}}=\frac{R}{\cos\frac{C}{2}}.$$

Therefore

$$P:Q:R=\cos\frac{A}{2}:\cos\frac{B}{2}:\cos\frac{C}{2}.$$

Option C which is: $$\cos\frac{A}{2} : \cos\frac{B}{2} : \cos\frac{C}{2}$$