Question

Differential coefficient of $tan^{-1} \frac{2x}{1-x^{2}}$ with respect to $sin^{-1} \frac{2x}{1+x^{2}}$ will be

• A. $1$
• B. $-1$
• C. $- 1/2$
• D. $x$

Answer 1

Answer: (A)

$1$

Answer 2

Let $u=tan^{-1} \frac{2x}{1-x^{2}} \,....\left(i\right)$ and $v=sin^{-1} \frac{2x}{1+x^{2}} \,....\left(ii\right)$ In equation $\left(i\right)$ put, $x = tan \theta$ $\therefore u=tan^{-1}\left[\frac{2\,tan\,\theta}{1-tan^{2}\,\theta}\right]=tan^{-1}\left(tan\,2\,\theta\right)$ $\Rightarrow u=2\,\theta \Rightarrow \frac{du}{d\theta }=2\,....\left(a\right)$ In equation $\left(ii\right)$, put $x = tan\,\theta$ $\therefore v=sin^{-1}\left[\frac{2\,tan\,\theta }{1+tan^{2}\,\theta }\right]=sin^{-1}\left(sin\,2\theta \right)$ $\Rightarrow v=2\theta \Rightarrow \frac{dv}{d\theta}=2 \,.....\left(b\right)$ From equations $\left(a\right)$ and $\left(b\right),$ $\frac{du}{dv}=\frac{du}{d\theta} \times\frac{d\theta}{dv}=2\times\frac{1}{2}=1$ $\therefore$ Required differential coefficient will be 1.