If y = ^ -1 x 1+ x then d y d x = ? - Vidyadip

Question

If $y =\tan ^{-1} \frac{\cos x}{1+\sin x}$ then $\frac{d y}{d x}=$ ?

✓

Answer

$(d) \frac{-1}{2}$
Explanation : Given that $y=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$
Using $\cos x=\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}, \sin x$
$=2 \sin \frac{x}{2} \cos \frac{x}{2}$ and $\cos ^2 \theta+\sin ^2 \theta=1$
Therefore,
$y=\tan ^{-1}\left(\frac{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}}{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}\right)=\tan ^{-1}\left(\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}\right)$
$\Rightarrow y =\tan ^{-1} \frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}$
Dividing by $\cos \frac{x}{2}$ in numerator and denominator, we get
$y=\tan ^{-1} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}$
Using $\tan \left(\frac{\pi}{4}-x\right)=\frac{1-\tan x}{1+\tan x}$, we obtain
$y=\tan ^{-1} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right)$
$=\frac{\pi}{4}-\frac{x}{2}$
Differentiating with respect to $x,$ we
$\frac{d y}{d x}=-\frac{1}{2}$

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