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

Hint:

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

Answer 1

The Correct Option is B

In this problem, we need to find the value of the expression:

\(\dfrac{\sqrt{3}\cosec 20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}\)

Let's solve it step-by-step:

1. Firstly, simplify the expression \( \sqrt{3}\cosec 20^\circ - \sec 20^\circ \).
2. Recall the trigonometric identities:
   • \(\cosec \theta = \dfrac{1}{\sin \theta}\)
   • \(\sec \theta = \dfrac{1}{\cos \theta}\)
3. So, the expression becomes: \(\sqrt{3}\dfrac{1}{\sin 20^\circ} - \dfrac{1}{\cos 20^\circ}\)
4. Combine the terms into a single fraction: \(\dfrac{\sqrt{3}\cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ}\)
5. The denominator simplifies to: \(\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ\)
6. Next, note:
   • \(\cos 60^\circ = \dfrac{1}{2}\)
7. The denominator can be rewritten as: \(\dfrac{1}{2}\cos 20^\circ \cos 40^\circ \cos 80^\circ\)
8. Using the identity for the product of cosines:
   • \(\cos 20^\circ \cos 40^\circ \cos 80^\circ = \dfrac{1}{8}\)
9. Thus, the complete denominator is: \(\dfrac{1}{16}\)
10. Now, the main expression becomes: \(\dfrac{\sqrt{3}\cos 20^\circ - \sin 20^\circ}{\dfrac{1}{16}}\)
11. Simplifying, \(16(\sqrt{3}\cos 20^\circ - \sin 20^\circ)\)
12. This expression simplifies to 64, matching the given correct option.

Therefore, the value of the expression is 64.