Compound Angle Formulas

We have, 

$$\operatorname{cosec}10°-\sqrt{3}\sec10°\ =\dfrac{1}{\sin10^{\circ\ }}-\dfrac{\sqrt{\ 3}}{\cos10^{\circ\ }}$$

$$=\dfrac{\cos10^{\circ\ }-\sqrt{\ 3}\sin10^{\circ\ }}{\sin10^{\circ\ }\cos10^{\circ\ }}$$

$$=2\times\ 2\times\ \dfrac{\frac{1}{2}\left(\cos10^{\circ\ }-\sqrt{\ 3}\sin10^{\circ\ }\right)}{2\times\ \sin10^{\circ\ }\cos10^{\circ\ }}$$

$$=4\times\ \dfrac{\frac{1}{2}\cos10^{\circ\ }-\dfrac{\sqrt{\ 3}}{2}\sin10^{\circ\ }}{2\times\sin10^{\circ\ }\cos10^{\circ\ }}$$

$$=4\times\ \dfrac{\sin30^{\circ\ }\cos10^{\circ\ }-\cos30^{\circ\ }\sin10^{\circ\ }}{2\sin10^{\circ\ }\cos10^{\circ\ }}$$

$$=4\times\ \dfrac{\sin\left(30-10\right)^{\circ\ }}{\sin\left(2\times\ 10\right)^{\circ\ }}=4\times\ \dfrac{\sin20^{\circ\ }}{\sin20^{\circ\ }}=4$$

The correct answer is Option A (4).

We know that - 
$$\sin(2A)=2\sin A\cos A$$
$$\sin(A-B)=\sin A\cos B-\cos A\sin B$$
$$\cos60^o=\dfrac{1}{2}$$

Thus, the given equation will be - 

$$=\dfrac{\sqrt{3} \cosec20^{\circ}-\sec20^{\circ}}{\cos20^{\circ}\cos40^{\circ}\cos60^{\circ}\cos80^{\circ}}$$

$$=\dfrac{\frac{\sqrt{3}}{\sin20^o}-\frac{1}{\cos20^o}}{\cos20^o\cos40^o\cos60^o\cos80^o}$$

$$=\dfrac{\frac{\sqrt{3}\cos20^o-\sin20^o}{\sin20^o\cos20^o}}{\cos20^o\cos40^o\cos60^o\cos80^o}$$

$$=\dfrac{\sqrt{3}\cos20^o-\sin20^o}{\sin20^o\cos20^o\cos20^o\cos40^o\cos60^o\cos80^o}$$

$$=\dfrac{\sqrt{3}\cos20^o-\sin20^o}{\frac{\sin40^o}{2}\cos20^o\cos40^o\cos60^o\cos80^o}$$

$$=\dfrac{2\left(\sqrt{3}\cos20^o-\sin20^o\right)}{\sin40^o\cos40^o\cos20^o\cos60^o\cos80^o}$$

$$=\dfrac{2\left(\sqrt{3}\cos20^o-\sin20^o\right)}{\frac{\sin80^o}{2}\cos20^o\cos60^o\cos80^o}$$

$$=\dfrac{2\times2\left(\sqrt{3}\cos20^o-\sin20^o\right)}{\sin80^o\cos80^o\cos20^o\times\frac{1}{2}}$$

$$=\dfrac{4\left(\sqrt{3}\cos20^o-\sin20^o\right)}{\frac{\sin160^o}{2}\cos20^o\frac{1}{2}}$$

$$=\dfrac{2^4\left(\sqrt{3}\cos20^o-\sin20^o\right)}{\sin160^o\cos20^o}$$

$$=\dfrac{2^4\left(\sqrt{3}\cos20^o-\sin20^o\right)}{\sin20^o\cos20^o}$$

$$=\dfrac{2^4 \cdot 2\left[\sin60^o\cos20^o-\cos60^o\sin20^o\right]}{\sin20^o\cos20^o}$$

$$=\dfrac{2^5\left[\sin60^o\cos20^o-\cos60^o\sin20^o\right]}{\frac{\sin40^o}{2}}$$

$$=\dfrac{2\times2\times2\times2\times2\times2\left[\sin40^o\right]}{\sin40^o}$$

$$=64$$

Step 1: Grouping and Converting to Sine and Cosine

We start by bringing the single tangent term to the denominator on the Left Hand Side to group our variables logically.

$$\frac{\tan(x + 100^\circ)}{\tan x} = \tan(x + 50^\circ) \tan(x - 50^\circ)$$

Next, we break everything down into sines and cosines using the fundamental identity $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$.

For the Left Hand Side, we have a fraction divided by a fraction. When dividing by a fraction, we multiply by its reciprocal.

$$\text{Left Side} = \frac{\frac{\sin(x+100^\circ)}{\cos(x+100^\circ)}}{\frac{\sin x}{\cos x}}$$
$$\text{Left Side} = \frac{\sin(x+100^\circ)}{\cos(x+100^\circ)} \times \frac{\cos x}{\sin x} = \frac{\sin(x+100^\circ) \cos x}{\cos(x+100^\circ) \sin x}$$

For the Right Hand Side, we simply multiply two tangent fractions straight across the numerators and denominators.

$$\text{Right Side} = \left( \frac{\sin(x+50^\circ)}{\cos(x+50^\circ)} \right) \left( \frac{\sin(x-50^\circ)}{\cos(x-50^\circ)} \right) = \frac{\sin(x+50^\circ) \sin(x-50^\circ)}{\cos(x+50^\circ) \cos(x-50^\circ)}$$

Equating our newly expanded sides gives us our working foundation:

$$\frac{\sin(x+100^\circ) \cos x}{\cos(x+100^\circ) \sin x} = \frac{\sin(x+50^\circ) \sin(x-50^\circ)}{\cos(x+50^\circ) \cos(x-50^\circ)}$$

Step 2: Applying Componendo and Dividendo

If $$\frac{a}{b} = \frac{c}{d}$$, we can instantly rewrite it as $$\frac{a+b}{a-b} = \frac{c+d}{c-d}$$. This allows us to rapidly build standard trigonometric identities.

Processing the Left Hand Side:

The new numerator becomes $$a+b$$:

$$\sin(x+100^\circ) \cos x + \cos(x+100^\circ) \sin x = \sin(x+100^\circ + x) = \sin(2x + 100^\circ)$$

The new denominator becomes $$a-b$$:

$$\sin(x+100^\circ) \cos x - \cos(x+100^\circ) \sin x = \sin(x+100^\circ - x) = \sin 100^\circ$$

Processing the Right Hand Side:

The new numerator becomes $$c+d$$:

$$\sin(x+50^\circ) \sin(x-50^\circ) + \cos(x+50^\circ) \cos(x-50^\circ) = \cos((x+50^\circ) - (x-50^\circ)) = \cos 100^\circ$$

The new denominator becomes $$c-d$$:

$$\sin(x+50^\circ) \sin(x-50^\circ) - \cos(x+50^\circ) \cos(x-50^\circ)$$
$$= -[ \cos(x+50^\circ) \cos(x-50^\circ) - \sin(x+50^\circ) \sin(x-50^\circ) ] = -\cos(x+50^\circ + x-50^\circ) = -\cos 2x$$

Putting the new simplified fractions together, our massive equation shrinks down to:

$$\frac{\sin(2x + 100^\circ)}{\sin 100^\circ} = \frac{\cos 100^\circ}{-\cos 2x}$$

Step 3: Cross Multiplication and Double Angle Identities

Cross multiply to remove the fractions completely:

$$-\sin(2x + 100^\circ) \cos 2x = \sin 100^\circ \cos 100^\circ$$

Multiply both sides of the equation by 2. This perfectly sets up the product to sum formula $$2\sin A \cos B = \sin(A+B) + \sin(A-B)$$ on the left, and the double angle formula $$2\sin \theta \cos \theta = \sin 2\theta$$ on the right.

$$-2\sin(2x + 100^\circ) \cos 2x = 2\sin 100^\circ \cos 100^\circ$$
$$-[ \sin(2x + 100^\circ + 2x) + \sin(2x + 100^\circ - 2x) ] = \sin 200^\circ$$
$$-[ \sin(4x + 100^\circ) + \sin 100^\circ ] = \sin 200^\circ$$

Move $$\sin 100^\circ$$ to the other side to isolate our $$x$$ variable:

$$-\sin(4x + 100^\circ) = \sin 200^\circ + \sin 100^\circ$$

Step 4: Sum to Product Conversion

We simplify the right side using the formula $$\sin C + \sin D = 2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$$:

$$\sin 200^\circ + \sin 100^\circ = 2\sin\left(\frac{300^\circ}{2}\right)\cos\left(\frac{100^\circ}{2}\right) = 2\sin 150^\circ \cos 50^\circ$$

Because $$\sin 150^\circ = \frac{1}{2}$$, this entire term collapses cleanly into just $$\cos 50^\circ$$.

$$-\sin(4x + 100^\circ) = \cos 50^\circ$$
$$\sin(4x + 100^\circ) = -\cos 50^\circ$$

Step 5: Solving for the Final Angles

To solve the equation, we need matching trigonometric functions. We must convert the negative cosine into a sine function.

$$-\cos 50^\circ = -\sin 40^\circ = \sin(-40^\circ)$$

This gives us our final trigonometric equation to solve:

$$\sin(4x + 100^\circ) = \sin(-40^\circ)$$

We plug this into the standard general solution formula $$\theta = n(180^\circ) + (-1)^n \alpha$$:

$$4x + 100^\circ = n(180^\circ) + (-1)^n (-40^\circ)$$

Now we test consecutive integer values for $$n$$ to discover the specific solutions that fit into the given interval between $$0^\circ$$ and $$180^\circ$$.

• For n = 1:

  $$4x + 100^\circ = 180^\circ - (-40^\circ) = 220^\circ$$
  $$4x = 120^\circ \implies x = 30^\circ$$

• For n = 2:

  $$4x + 100^\circ = 360^\circ + (-40^\circ) = 320^\circ$$
  $$4x = 220^\circ \implies x = 55^\circ$$

• For n = 3:

  $$4x + 100^\circ = 540^\circ - (-40^\circ) = 580^\circ$$
  $$4x = 480^\circ \implies x = 120^\circ$$

• For n = 4:

  $$4x + 100^\circ = 720^\circ + (-40^\circ) = 680^\circ$$
  $$4x = 580^\circ \implies x = 145^\circ$$

If we try $$n = 0$$, we get $$x = -35^\circ$$, and if we try $$n = 5$$, we get $$x = 210^\circ$$. Both of these fall strictly outside the required domain boundaries.

The valid angles are exactly $$30^\circ$$, $$55^\circ$$, $$120^\circ$$, and $$145^\circ$$.

The final number of elements in the set is 4.