Q84 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics

If $y = \cos^2\left[\cot^{-1}\left(\sqrt{\dfrac{1 - x}{1 + x}}\right)\right]$ then $\dfrac{dy}{dx} = $ ........

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

Solution

Let $x = \cos \theta$.

Substituting this in the given expression:

$\sqrt{\dfrac{1 - x}{1 + x}} = \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta}} = \sqrt{\dfrac{2 \sin^2(\theta/2)}{2 \cos^2(\theta/2)}} = \tan\left(\dfrac{\theta}{2}\right)$

The expression for $y$ becomes:

$y = \cos^2\left[\cot^{-1}\left(\tan\left(\dfrac{\theta}{2}\right)\right)\right]$

Using the identity $\tan\left(\dfrac{\theta}{2}\right) = \cot\left(\dfrac{\pi}{2} - \dfrac{\theta}{2}\right)$:

$y = \cos^2\left[\cot^{-1}\left(\cot\left(\dfrac{\pi}{2} - \dfrac{\theta}{2}\right)\right)\right]$

$y = \cos^2\left(\dfrac{\pi}{2} - \dfrac{\theta}{2}\right)$

$y = \sin^2\left(\dfrac{\theta}{2}\right)$

Using the half-angle formula $\sin^2\left(\dfrac{\theta}{2}\right) = \dfrac{1 - \cos \theta}{2}$:

$y = \dfrac{1 - x}{2}$

Differentiating with respect to $x$:

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

Answer: $\dfrac{-1}{2}$

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

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