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MHTCET 13 Apr 2026 Shift 2 Paper

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Question : 121 of 150

Marks: +1, -0

\frac{d}{d x}\left[\sin ^{2}\left\{\cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right\}\right]=\ldots

-1
$\frac{1}{2}$
$\frac{-1}{2}$
1

Solution:

Concept:

Simplify the given composite trigonometric function before differentiating.

Explanation:

Let

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

.

Put

\theta = \cot^{-1}\sqrt{\frac{1-x}{1+x}}

.

Then

\cot\theta = \sqrt{\frac{1-x}{1+x}}

.

Squaring both sides,

\cot^2\theta = \frac{1-x}{1+x}

.

Using

1+\cot^2\theta = \csc^2\theta

, we get

\sin^2\theta = \frac{1}{1+\cot^2\theta}

.

Substitute the value:

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

.

Hence

y = \frac{1+x}{2}

.

Differentiating with respect to

x

,

\frac{dy}{dx} = \frac{1}{2}

.

Answer:

\frac{1}{2}

, which is option B.

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