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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation3 Marks
Question
Differentiate the following w. r. t. x.$\cot ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$
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Answer

Let $y=\cot ^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right)$
$=\tan ^{-1}\left(\frac{\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]^2}{\cos ^2\left(\frac{x}{2}\right)-\sin ^2\left(\frac{x}{2}\right)}\right)=\tan ^{-1}\left(\frac{\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]^2}{\left[\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right]\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]}\right)$
$=\tan ^{-1}\left(\frac{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}\right)=\tan ^{-1}\left(\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\right) \ldots$. Divide Numerator \& Denominator by $\cos \left(\frac{x}{2}\right)$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]$
$\therefore y=\frac{\pi}{4}+\frac{x}{2}$
Differentiate $w . r . t . x$.
$
\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}+\frac{x}{2}\right)=0+\frac{1}{2} \quad \therefore \frac{d y}{d x}=\frac{1}{2}
$

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