Q67 / 150 · MHT CET 2021 (20 Sep Shift 1) · Mathematics

If $y=\log \tan \left(\frac{x}{2}\right)+\sin ^1(\cos x)$, then $\frac{d y}{d x}=$

  • A
    $\operatorname{cosec} x$
  • B
    $\sin x+1$
  • C
    x
  • Correct answer
    D
    $\operatorname{cosec} x-1$

Solution

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

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

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