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

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

Answers (1)

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}

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