The value of 3 20^ - 20^ 20^ 40^ 60^ 80^ is equal to — Solution | JEE Mathematics

MCQMediumJEE 2026Trigonometric Ratios & Identities

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

Answer

Correct answer:B

Standard Method

Given: Evaluate

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

Find: The numerical value of $E$ .

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)

we get

=\frac{2\cos 50^\circ}{\sin 20^\circ\cos 20^\circ}

Step 2: Simplifying the denominator.

\cos 60^\circ=\frac{1}{2}

Also,

\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.

E=\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,

E=64

Therefore, the correct option is B.

Using the identity $\sqrt{3}\cos \theta-\sin \theta$ incorrectly. The correct reduction is $2\cos(\theta+30^\circ)$ , not $2\sin(\theta+30^\circ)$ . Rewrite the numerator carefully before combining with the denominator.
Forgetting the factor from $\cos 60^\circ$ . Since $\cos 60^\circ=\frac{1}{2}$ , the denominator product becomes $\frac{\sin 80^\circ}{8}$ , not $\frac{\sin 80^\circ}{4}$ . Track this factor separately.
Applying the product identity for $\cos 20^\circ\cos 40^\circ\cos 80^\circ$ wrongly. A common error is to replace it with an incorrect sine product. Use the standard identity exactly as stated in the solution.

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