Q84 / 150 · MHT CET 2026 (13 April Shift 2) · Mathematics

$\dfrac{d}{dx}\left[\sin^2\left\{\cot^{-1}\sqrt{\dfrac{1-x}{1+x}}\right\}\right] =$ ...

  • A
    $-1$
  • Correct answer
    B
    $\dfrac{1}{2}$
  • C
    $\dfrac{-1}{2}$
  • D
    $1$

Solution

Let $y = \sin^2\left\{\cot^{-1}\sqrt{\dfrac{1-x}{1+x}}\right\}$

Let $\cot^{-1}\sqrt{\dfrac{1-x}{1+x}} = \theta$

$\Rightarrow \cot \theta = \sqrt{\dfrac{1-x}{1+x}}$

$\Rightarrow \cot^2 \theta = \dfrac{1-x}{1+x}$

We know that $\sin^2 \theta = \dfrac{1}{\csc^2 \theta} = \dfrac{1}{1 + \cot^2 \theta}$

Substituting the value of $\cot^2 \theta$:

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

Therefore, $y = \dfrac{1+x}{2}$

Differentiating with respect to $x$:

$\dfrac{dy}{dx} = \dfrac{d}{dx}\left(\dfrac{1+x}{2}\right) = \dfrac{1}{2}$

Hence, the correct answer is option: 'B'.

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