(A) 32

(B) 64

(C) 12

(D) 16

Correct Answer: 64

Explanation

First simplify the numerator:

$$ \sqrt{3}\csc20^\circ-\sec20^\circ =\frac{\sqrt{3}}{\sin20^\circ}-\frac{1}{\cos20^\circ} $$

$$ =\frac{\sqrt{3}\cos20^\circ-\sin20^\circ}{\sin20^\circ\cos20^\circ} $$

Using identity:

$$ \sqrt{3}\cos\theta-\sin\theta =2\cos\left(\theta+30^\circ\right) $$

For $\theta=20^\circ$:

$$ \sqrt{3}\cos20^\circ-\sin20^\circ =2\cos50^\circ $$

Hence numerator becomes:

$$ \frac{2\cos50^\circ}{\sin20^\circ\cos20^\circ} $$

Now denominator:

$$ \cos20^\circ\cos40^\circ\cos60^\circ\cos80^\circ =\frac12\cos20^\circ\cos40^\circ\cos80^\circ $$

Using identity:

$$ \cos20^\circ\cos40^\circ\cos80^\circ=\frac18 $$

Thus denominator:

$$ =\frac12\times\frac18=\frac1{16} $$

Therefore the whole expression becomes:

$$ \frac{2\cos50^\circ}{\sin20^\circ\cos20^\circ}\div\frac1{16} =32\cdot\frac{\cos50^\circ}{\sin20^\circ\cos20^\circ} $$

Since $\cos50^\circ=\sin40^\circ$,

$$ =32\cdot\frac{\sin40^\circ}{\sin20^\circ\cos20^\circ} $$

Using $\sin40^\circ=2\sin20^\circ\cos20^\circ$:

$$ =32\times2=64 $$

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