Q104 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Mathematics

$\frac{d}{d x}\left[\operatorname{cosech}^{-1}(\tan 2 x)\right]=$

  • A
    $2|\sec 2 x|$
  • B
    $\cos 2 x$
  • Correct answer
    C
    $-2|\operatorname{cosec} 2 x|$
  • D
    $\sin 2 x$

Solution

We have, $\frac{d}{d x}\left(\operatorname{cosec~}^{-1}(\tan 2 x)\right.$ $\begin{aligned} & =\frac{-1}{|\tan 2 x|} \times \frac{1}{\sqrt{1+\tan ^2 2 x}} \times 2 \sec ^2 2 x \\ & {\left[\frac{d}{d x} \operatorname{cosec~h}^{-1} x=\frac{-1}{|x| \sqrt{1+x^2}}\right]} \\ & =-2|\operatorname{cosec} 2 x| \end{aligned}$

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

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