Question

Assertion :Derivative of ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ w.r.t ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is $1,$ for $0<x<1$. Reason: ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ for $-1\le x\le 1$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

${\sin }^{-1}\frac{2x}{1+x^2}=2{\tan }^{-1}x$ for $-1\le x\le 1$

and ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\{\begin{array}{c}\begin{array}{cc}2{\tan }^{-1}x,& x\ge 0\end{array}\\ \begin{array}{cc}-2{\tan }^{-1}x,& x<0\end{array}\end{array}$

$\therefore$ For $0<x<1,{\sin }^{-1}\frac{2x}{1+x^2}={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

$\therefore$ Derivative of ${\sin }^{-1}\frac{2x}{1+x^2}$ w.r.t ${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is $1$

$\therefore$ Assertion is true but reason is false.