Let $a$ solution $y=y(x)$ of the differential equation $x \sqrt{x^2-1} d y-y \sqrt{y^2-1} d x=0$ satisfy $y(2)=\frac{2}{\sqrt{3}}$

Statement-1: $y(x)=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)$

Statement-2: $y(x)$ is given by $\frac{1}{y}=\frac{2 \sqrt{3}}{x}-\sqrt{1-\frac{1}{x^2}}$

Statement-1 is True, Statement-2 is False.

We have,

$x \sqrt{x^2-1} d y=y \sqrt{y^2-1} d x$

$\Rightarrow \frac{1}{y \sqrt{y^2-1}} d y=\frac{1}{x \sqrt{x^2-1}} d x$

$\Rightarrow \int \frac{1}{y \sqrt{y^2-1}} d y=\int \frac{1}{x \sqrt{x^2-1}} d x$

$\Rightarrow \sec ^{-1} y=\sec ^{-1} x+C$        .........(i)

It is given that $y=\frac{2}{\sqrt{3}}$ when $x=2$

∴  $\sec ^{-1} \frac{2}{\sqrt{3}}=\sec ^{-1} 2+C \Rightarrow \frac{\pi}{6}=\frac{\pi}{3}+C \Rightarrow C=-\frac{\pi}{6}$

Putting $C=-\frac{\pi}{6}$ in (i), we get

$\sec ^{-1} y=\sec ^{-1} x-\frac{\pi}{6}$         ........(ii)

$\Rightarrow y=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)$

So, statement- 1 is true.

From (ii), we have

$\cos ^{-1}\left(\frac{1}{y}\right)=\cos ^{-1}\left(\frac{1}{x}\right)-\frac{\pi}{6}$

$\Rightarrow \frac{1}{y}=\cos \left\{\cos ^{-1}\left(\frac{1}{x}\right)-\frac{\pi}{6}\right\}$

$\Rightarrow \frac{1}{y}=\cos \left\{\cos ^{-1}\left(\frac{1}{x}\right)\right\} \cos \left(\frac{\pi}{6}\right)+\sin \left(\cos ^{-1} \frac{1}{x}\right) \sin \frac{\pi}{6}$

$\Rightarrow \frac{1}{y}=\frac{\sqrt{3}}{2 x}+\frac{1}{2} \sqrt{1-\frac{1}{x^2}}$

So, statement-2 is false.