#### Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 14

Answer: $\frac{-x \sqrt{x^{2}-1}}{2}$

Hint: $\text { Let } u=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right), v=\sec ^{-1} x$

Given: $\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right) \text { w.r.t } \sec ^{-1} x$

Explanation:

$\text { Let } u=\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$

\begin{aligned} &v=\sec ^{-1} x \\\\ &u=\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) \end{aligned}

$\left[\begin{array}{l} \cos 2 x=\cos ^{2} x-\sin ^{2} x \\ \sin 2 x=2 \sin x \cos x \\ 1=\cos ^{2} x+\sin ^{2} x \end{array}\right]$

$u=\tan ^{-1}\left(\frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}\right)$

$u=\tan ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right)$

Divide numerator and denominator by $\cos \frac{x}{2}$

$u=\tan ^{-1}\left(\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right)$

$u=\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}-\tan \frac{x}{2}}{1+\tan \frac{\pi}{4} \tan \frac{x}{2}}\right)$

\begin{aligned} &u=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right) \\\\ &u=\frac{\pi}{4}-\frac{x}{2} \\\\ &\frac{d u}{d x}=\frac{-1}{2} \end{aligned}

\begin{aligned} &v=\sec ^{-1} x \\\\ &\frac{d v}{d x}=\frac{1}{x \sqrt{x^{2}-1}} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-1}{2}}{\frac{1}{x\sqrt{x^{2}-1}}}=\frac{-x \sqrt{x^{2}-1}}{2} \end{aligned}