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MHTCET 16 May 2024 Shift 1 Paper

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Question : 110 of 150

Marks: +1, -0

If

\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}

, then the value of

x

is

4
5
1
3

Solution:

Concept:

Use the identity

\operatorname{cosec}^{-1} a = \sin^{-1}\left(\frac{1}{a}\right)

and

\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}

.

Explanation:

Given:

\sin^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}

.

Since

\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\sin^{-1}\left(\frac{4}{5}\right)

, we get:

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

.

Therefore,

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

.

Now,

\cos^{-1}\left(\frac{4}{5}\right)=\sin^{-1}\left(\frac{3}{5}\right)

because

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

.

So,

\sin^{-1}\left(\frac{x}{5}\right)=\sin^{-1}\left(\frac{3}{5}\right)

, which gives

\frac{x}{5}=\frac{3}{5}

.

Hence,

x=3

.

Answer:

x=3

, i.e., option D.

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