Question

The value of \(\dfrac{\sqrt{3}\cosec 20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}\) is equal to 
 

• A. 32
• B. 64
• C. 12
• D. 16
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Hint:

Always look for standard trigonometric product identities when angles differ by $20^\circ$.

Answer 1

The Correct Option is B

Step 1: Simplifying the numerator.
\[ \sqrt{3}\cosec 20^\circ - \sec 20^\circ = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ} \] Taking LCM, \[ = \frac{\sqrt{3}\cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ} \] Using identity \[ \sqrt{3}\cos \theta - \sin \theta = 2\cos(\theta + 30^\circ) \] \[ = \frac{2\cos 50^\circ}{\sin 20^\circ \cos 20^\circ} \] Step 2: Simplifying the denominator.
\[ \cos 60^\circ = \frac{1}{2} \] \[ \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{\sin 80^\circ}{4} \] Thus, \[ \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{\sin 80^\circ}{8} \] Step 3: Final evaluation.
\[ \text{Expression} = \frac{2\cos 50^\circ}{\sin 20^\circ \cos 20^\circ} \times \frac{8}{\sin 80^\circ} \] Using $\sin 80^\circ = \cos 10^\circ$ and simplification, \[ = 64 \]