Differential coefficient of $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$ w.r.t. $\sin ^{-1} \frac{2 x}{1+x^2}$ is equal to :
Answer & explanation
Correct answer: option 3
Put y = $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$ and $z=\sin ^{-1} \frac{2 x}{1+x^2}$
Put x = $\tan \theta \Rightarrow \theta=\tan ^{-1} x$
∴ $y=\tan ^{-1}(\tan 2 \theta)=2 \theta, z=\sin ^{-1}(\sin 2 \theta)=2 \theta$
$\frac{d y}{d \theta}=2, \frac{d z}{d \theta}=2 \Rightarrow \frac{d y}{d z}=1$
Hence (3) is correct answer.