Choose the correct alternative in the following:

We are given that

$U=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), V=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$

$\frac{\mathrm{dU}}{\mathrm{dV}}=?$

Now, we know

$\frac{\mathrm{dU}}{\mathrm{dV}}=\frac{\frac{\mathrm{dU}}{\mathrm{dx}}}{\frac{\mathrm{dV}}{\mathrm{dx}}}$

Now,

$\frac{d U}{d x}=\frac{d}{d x} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

Put $x=\tan \theta$

$\theta=\tan ^{-1} x \cdots(1)$

$\Rightarrow \frac{\mathrm{dU}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$

$=\frac{\mathrm{d}}{\mathrm{dx}} \sin ^{-1}(\sin 2 \theta) \because \frac{2 \tan \theta}{1+\tan ^{2} \theta}=\sin 2 \theta$

$=\frac{\mathrm{d}}{\mathrm{dx}} 2 \theta$

$=\frac{\mathrm{d}}{\mathrm{dx}} 2 \tan ^{-1} \mathrm{x}=\frac{2}{1+\mathrm{x}^{2}}$

Again

$\frac{d V}{d x}=\frac{d}{d x} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$

Put $x=\tan \theta$

$\theta=\tan ^{-1} x \ldots(1)$

$\Rightarrow \frac{\mathrm{dV}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} \tan ^{-1}\left(\frac{2 \tan \theta}{1-\tan ^{2} \theta}\right)$

$=\frac{\mathrm{d}}{\mathrm{dx}} \tan ^{-1}(\tan 2 \theta) \because \frac{2 \tan \theta}{1-\tan ^{2} \theta}=\tan 2 \theta$

$=\frac{\mathrm{d}}{\mathrm{dx}} 2 \theta$

$=\frac{d}{d x} 2 \tan ^{-1} x=\frac{2}{1+x^{2}} \cdots$ From (1)

Now, $\frac{\mathrm{dU}}{\mathrm{dV}}=\frac{\frac{\mathrm{du}}{\mathrm{dx}}}{\frac{\mathrm{dv}}{\mathrm{dx}}}=\frac{\frac{2}{1+\mathrm{x}^{2}}}{\frac{2}{1+\mathrm{x}^{2}}}=1$

$\therefore \frac{\mathrm{dU}}{\mathrm{dV}}=1=(\mathrm{D})$