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MHTCET 15 May 2024 Shift 2 Paper

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Question : 124 of 150

Marks: +1, -0

If

\cos ^{-1} x=\alpha(0 < x < 1)

\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)+\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)=\frac{2 \pi}{3}

\alpha

is

$\frac{\pi}{2}$
$\frac{\pi}{6}$
$\frac{\pi}{3}$
$\frac{\pi}{4}$

Solution:

Concept:

Use substitution

x = \cos \alpha

to simplify the inverse trigonometric expressions.

Explanation:

Given

\cos^{-1} x = \alpha

and

0 < x < 1

, we get

0 < \alpha < \frac{\pi}{2}

.

So

x = \cos \alpha

and

\sqrt{1-x^2} = \sin \alpha

.

Substitute in the given equation:

\sin^{-1}\left(2\cos\alpha \sin\alpha\right) + \sec^{-1}\left(\frac{1}{2\cos^2\alpha - 1}\right)=\frac{2\pi}{3}

Using

2\cos\alpha\sin\alpha = \sin 2\alpha

and

2\cos^2\alpha - 1 = \cos 2\alpha

:

\sin^{-1}(\sin 2\alpha) + \sec^{-1}(\sec 2\alpha) = \frac{2\pi}{3}

For

0 < \alpha < \frac{\pi}{2}

, we have

0 < 2\alpha < \pi

.

The valid branch here gives both terms equal to

2\alpha

.

Therefore:

2\alpha + 2\alpha = \frac{2\pi}{3}

4\alpha = \frac{2\pi}{3}

\alpha = \frac{\pi}{6}

Answer:

\alpha = \frac{\pi}{6}

, i.e. option B.

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