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GUJCET 2025 Math Paper

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Question : 12 of 40

Marks: +1, -0

\tan^{-1} \left[ \frac{\sqrt{2}}{\sqrt{3}} \cos \left( 5 \sin^{-1} \frac{1}{\sqrt{2}} \right) \right] = \dots

$-\pi/3$
$\pi/3$
$-\pi/6$
$\pi/6$

Solution:

Concept:

Simplify the inverse trigonometric expression step by step.

Explanation:

Given

\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}

.

So,

5\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)=5\cdot\frac{\pi}{4}=\frac{5\pi}{4}

.

Now,

\cos\left(\frac{5\pi}{4}\right)=\cos\left(\pi+\frac{\pi}{4}\right)=-\cos\left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}

.

Substitute this value:

\frac{\sqrt{2}}{\sqrt{3}}\cos\left(\frac{5\pi}{4}\right)=\frac{\sqrt{2}}{\sqrt{3}}\cdot\left(-\frac{1}{\sqrt{2}}\right)=-\frac{1}{\sqrt{3}}

.

Thus the expression becomes

\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)

.

Since

\tan^{-1}(-x)=-\tan^{-1}(x)

and

\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}

,

the final value is

-\frac{\pi}{6}

.

Answer:

\boxed{-\frac{\pi}{6}}

Option C is correct.

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