Choose the correct alternative in the following:

Solution:

$\sin ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=\log a$

$\frac{x^{2}-y^{2}}{x^{2}+y^{2}}=\sin (\log a)$

Put $y=x \tan \theta$

$\theta=\tan ^{-1}\left(\frac{y}{x}\right) \cdots(1)$

$\Rightarrow \frac{x^{2}-x^{2} \tan ^{2} \theta}{x^{2}+x^{2} \tan ^{2} \theta}=\sin (\log a)$

$\Rightarrow \frac{x^{2}-x^{2} \tan ^{2} \theta}{x^{2}+x^{2} \tan ^{2} \theta}=\sin (\log a)$

$\Rightarrow \frac{x^{2}\left(1-\tan ^{2} \theta\right)}{x^{2}\left(1+\tan ^{2} \theta\right)}=\sin (\log a)$

$\Rightarrow \cos 2 \theta=\sin (\log a) \because \frac{\left(1-\tan ^{2} \theta\right)}{\left(1+\tan ^{2} \theta\right)}=\cos 2 \theta$

$\Rightarrow 2 \theta=\cos ^{-1}[\sin (\log a)]$

$\Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)=\frac{1}{2} \cos ^{-1}[\sin (\log a)]$

Taking tan on both sides

$\Rightarrow \tan \left[\tan ^{-1}\left(\frac{y}{x}\right)\right]=\tan \left[\frac{1}{2} \cos ^{-1}[\sin (\log a)]\right]$

$\Rightarrow \frac{y}{x}=\tan \left[\frac{1}{2} \cos ^{-1}[\sin (\log a)]\right]$

Differentiating w.r.t $\mathrm{x}$ we get,

$\Rightarrow \frac{\frac{d y}{d x} \cdot x-y \cdot \frac{d x}{d x}}{x^{2}}=0 \because \tan \left[\frac{1}{2} \cos ^{-1}[\sin (\log a)]\right]$ is a constant

$\Rightarrow x \cdot \frac{d y}{d x}-y=0$

$\therefore \frac{d y}{d x}=\frac{y}{x}$