#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.8 question 9 maths textbook solution

Hint:   $\text { Let } u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$
Given:   $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \text { w.r.t } \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$

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Explanation:  $\text { Let } u=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

\begin{aligned} &v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\\\ &\text { Let } x=\tan \theta \\\\ &u=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right) \end{aligned}

\begin{aligned} &=\sin ^{-1}(\sin 2 \theta) \\\\ &0

\begin{aligned} &u=\sin ^{-1}(\sin 2 \theta)=2 \theta\; \; \; \; \; \; \quad 2 \theta \in\left(0, \frac{\pi}{2}\right) \\ &u=2 \tan ^{-1} x \end{aligned}

\begin{aligned} &\frac{d u}{d x}=\frac{2}{1+x^{2}} \\\\ &v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{aligned}

\begin{aligned} &v=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \\\\ &=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \\\\ &=\cos ^{-1}(\cos 2 \theta) \end{aligned}

\begin{aligned} &=2 \theta\; \; \; \; \; \; \; \; \quad 2 \theta \in\left(0, \frac{\pi}{2}\right) \\\\ &v=2 \tan ^{-1} x \\\\ &\frac{d v}{d x}=\frac{2}{1+x^{2}} \end{aligned}

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