Test Index

GUJCET 2026 Math Paper

© examsnet.com

Question : 12 of 40

Marks: +1, -0

\tan^{-1} [2 \cos (2 \sin^{-1} \frac{1}{2})] = \dots \dots \dots

$-\frac{\pi}{4}$
$\frac{\pi}{4}$
$\frac{3\pi}{4}$
$-\frac{3\pi}{4}$

Solution:

Concept:

Use standard principal values of inverse trigonometric functions.

Explanation:

Let

\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}

.

Then

2\sin^{-1}\left(\frac{1}{2}\right)=2\cdot\frac{\pi}{6}=\frac{\pi}{3}

.

So the expression becomes

\tan^{-1}\left[2\cos\left(\frac{\pi}{3}\right)\right]

.

Since

\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}

, we get

2\cos\left(\frac{\pi}{3}\right)=2\cdot\frac{1}{2}=1

.

Thus

\tan^{-1}(1)=\frac{\pi}{4}

, because

\tan\left(\frac{\pi}{4}\right)=1

.

Answer:

\frac{\pi}{4}

© examsnet.com

Go to Question:

Prev Question

Next Question