#### Explain solution RD Sharma class 12 chapter Differentiation exercise 10.8 question 12 maths

Hint:  $\text { : Let } u=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right), v=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$

Given:  $\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \text { w.r.t } \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$

$0

Explanation:

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

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

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

\begin{aligned} &=\cos ^{-1}(\cos 2 \theta)\\\\ &\text { Now, }\\\\ &0

\begin{aligned} &0<\theta<\frac{\pi}{4} \\\\ &0<2 \theta<\frac{\pi}{2} \\\\ &u=\tan ^{-1}(\tan 2 \theta) \end{aligned}

$\begin{array}{ll} =2 \theta & 2 \theta \in\left(0, \frac{\pi}{2}\right) \\\\ =2 \tan ^{-1} x & \left(\begin{array}{l} \tan \theta=x \\ \theta=\tan ^{-1} x \end{array}\right) \end{array}$

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

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

$\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$

$=\frac{\frac{2}{1+x^{2}}}{\frac{2}{1+x^{2}}}=1$