Question

The differential coefficient of ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ w.r.t ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ is

• A. $1$
• B. $-1$
• C. $0$
• D. None of these

Answer 1

The correct option is A

$1$

Explanation for the correct option :

Let $u={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ and $v={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$.

As we know that ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=2{\tan }^{-1}x$ and ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=2{\tan }^{-1}x$, we have

$u=2{\tan }^{-1}x$ and $v=2{\tan }^{-1}x$ i.e. $u=v$ i.e. ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$.

Thus the differential coefficient of ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ w.r.t. ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ must be $1$.

Hence, option (A) is the correct answer.