Question

$\frac{d}{dx}\left({\cos }^{-1}\frac{1-x^2}{1+x^2}\right)$

• A. $2.\frac{1}{1+x^2}$
• B. $2.\frac{1}{1-x^2}$
• C. $2.\frac{-1}{1+x^2}$
• D. $2.\frac{-1}{1-x^2}$

Answer 1

The correct option is A $2.\frac{1}{1+x^2}$

Let $y={\cos }^{-1}\frac{1-x^2}{1+x^2}$

put $x=\tan \theta$

$\Rightarrow y={\cos }^{-1}\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }={\cos }^{-1}\cos 2\theta =2\theta =2{\tan }^{-1}x$

$\therefore \frac{dy}{dx}=2.\frac{1}{1+x^2}$