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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]3 Marks
Question
Differentiate $\cot ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$ w.r. to $x$
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Answer

$ \text { Let } y =\cot ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$
$=\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right)$
$=\tan ^{-1}\left[\frac{\cos ^2\left(\frac{x}{2}\right)+\sin ^2\left(\frac{x}{2}\right)+2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{\cos ^2\left(\frac{x}{2}\right)-\sin ^2\left(\frac{x}{2}\right)}\right]$
$=\tan ^{-1}\left[\frac{\left\{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right\}^2}{\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]\left[\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right]}\right]$
$=\tan ^{-1}\left[\frac{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}\right]$
$=\tan ^{-1}\left[\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\right]$
$=\tan ^{-1}\left[\frac{\tan \left(\frac{\pi}{4}\right)+\tan \left(\frac{\pi}{2}\right)}{1-\tan \left(\frac{\pi}{4}\right) \tan \left(\frac{x}{2}\right)}\right]$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]$
$\therefore y=\frac{\pi}{4}+\frac{x}{2} $
Differentiating w. r. t. x, we get
$\frac{ d y}{ d x}=\frac{ d }{ d x}\left(\frac{\pi}{4}+\frac{x}{2}\right)=0+\frac{1}{2}=\frac{1}{2}$

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