Q82 / 150 · MHT CET 2021 (22 Sep Shift 2) · Mathematics

If $y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$, then $\frac{d y}{d x}=$

  • A
    $1$
  • B
    $\frac{3}{2}$
  • Correct answer
    C
    $\frac{1}{2}$
  • D
    $\frac{-1}{2}$

Solution

$\begin{aligned} & y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}} \\ & =\tan ^{-1} \sqrt{\frac{2 \cos ^2 \frac{x}{2}}{2 \sin ^2 \frac{x}{2}}}=\tan ^{-1}\left(\cot \frac{x}{2}\right)=\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right]=\frac{\pi}{2}-\frac{x}{2} \\ & \therefore \frac{d y}{d x}=0-\frac{1}{2}=\frac{-1}{2}\end{aligned}$

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

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