1. Given differential equation

$x\sqrt{x^2-1}\,dy-y\sqrt{y^2-1}\,dx=0$

Rewrite it as

$x\sqrt{x^2-1}\,\frac{dy}{dx}=y\sqrt{y^2-1}$

So,

$\frac{dy}{y\sqrt{y^2-1}}=\frac{dx}{x\sqrt{x^2-1}}$

This is separable.

2. Integrate both sides

We use the standard result:

$\int \frac{dt}{t\sqrt{t^2-1}}=\sec^{-1} t + C$

Hence,

$\sec^{-1} y = \sec^{-1} x + C$

or

$\sec^{-1} y - \sec^{-1} x = C$

3. Use the initial condition

Given

$y(2)=\frac{2}{\sqrt3}$

So,

$\sec^{-1}\left(\frac{2}{\sqrt3}\right)-\sec^{-1}(2)=C$

Now,

$\sec\left(\frac{\pi}{6}\right)=\frac{2}{\sqrt3} \quad \Rightarrow \quad \sec^{-1}\left(\frac{2}{\sqrt3}\right)=\frac{\pi}{6}$

and

$\sec\left(\frac{\pi}{3}\right)=2 \quad \Rightarrow \quad \sec^{-1}(2)=\frac{\pi}{3}$

Thus,

$C=\frac{\pi}{6}-\frac{\pi}{3}=-\frac{\pi}{6}$

Therefore,

$\sec^{-1} y = \sec^{-1} x - \frac{\pi}{6}$

Taking secant on both sides,

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

So STATEMENT-1 is true.

4. Now simplify STATEMENT-1 to check STATEMENT-2

Let

$\theta=\sec^{-1}x$

Then

$\sec\theta=x, \qquad \cos\theta=\frac1x, \qquad \sin\theta=\sqrt{1-\frac1{x^2}}$

Now,

$\frac1y=\cos\left(\theta-\frac{\pi}{6}\right)$

Using

$\cos(A-B)=\cos A\cos B+\sin A\sin B$

we get

$\frac1y=\cos\theta\cos\frac{\pi}{6}+\sin\theta\sin\frac{\pi}{6}$

$\frac1y=\frac1x\cdot \frac{\sqrt3}{2}+\sqrt{1-\frac1{x^2}}\cdot \frac12$

So,

$\frac1y=\frac{\sqrt3}{2x}+\frac12\sqrt{1-\frac1{x^2}}$

This is not equal to

$\frac1y=\frac{2\sqrt3}{x}-\sqrt{1-\frac1{x^2}}$

Hence STATEMENT-2 is false.

5. Evaluate options

• A: false
• B: false
• C: true
• D: false

Therefore, the correct option is

$\boxed{\text{C}}$

6. Comparison with stored answer

Stored correct answer: C

My derived answer: C

They agree.