Q79 / 150 · MHT CET 2011 · Mathematics

$y=\log \tan x / 2+\sin ^{-1}(\cos x)$, then $d y / d x$ is

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

Solution

$\begin{aligned} \frac{d y}{d x} &=\frac{1}{\tan x / 2} \sec ^{2} \frac{x}{2} \cdot \frac{1}{2}+\frac{1}{\sqrt{1-\cos ^{2} x}}(-\sin x) \\ &=\frac{1}{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}-1=\operatorname{cosec} x-1 \end{aligned}$

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

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