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MHTCET 16 May 2024 Shift 2 Paper

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Question : 107 of 150

Marks: +1, -0

If

\cos ^{-1}\left(\frac{12}{13}\right)+\sin ^{-1}\left(\frac{3}{5}\right)=\sin ^{-1} \mathrm{P}

, then the value of

P

is

$\frac{63}{65}$
$\frac{56}{65}$
$\frac{48}{65}$
$\frac{36}{65}$

Solution:

Concept:

Convert

\cos^{-1}

to

\sin^{-1}

and use the standard addition formula for inverse sine.

Explanation:

Given:

\cos^{-1}\left(\frac{12}{13}\right)+\sin^{-1}\left(\frac{3}{5}\right)=\sin^{-1}P

Let

\cos^{-1}\left(\frac{12}{13}\right)=\sin^{-1}\left(\frac{5}{13}\right)

because

\sin\theta=\sqrt{1-\cos^2\theta}

.

So the expression becomes

\sin^{-1}\left(\frac{3}{5}\right)+\sin^{-1}\left(\frac{5}{13}\right)

.

Use the formula:

\sin^{-1}x+\sin^{-1}y=\sin^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)

.

Here,

x=\frac{3}{5}

and

y=\frac{5}{13}

.

Compute:

\sqrt{1-y^2}=\sqrt{1-\frac{25}{169}}=\frac{12}{13}

And

\sqrt{1-x^2}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}

Therefore,

\sin^{-1}P=\sin^{-1}\left(\frac{3}{5}\times\frac{12}{13}+\frac{5}{13}\times\frac{4}{5}\right)

This simplifies to

\sin^{-1}\left(\frac{36}{65}+\frac{20}{65}\right)=\sin^{-1}\left(\frac{56}{65}\right)

Hence,

P=\frac{56}{65}

.

Answer:

P=\frac{56}{65}

Correct option: B.

\frac{56}{65}

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