Solution

Given,

$x-y\tan 35^\circ=\tan 25^\circ(y+x\tan 35^\circ)$

Let

$\tan 35^\circ=A$

and

$\tan 25^\circ=B$

Then,

$x-yA=B(y+xA)$

$x-yA=By+ABx$

$x-ABx=yA+By$

$x(1-AB)=y(A+B)$

So,

$\frac{x}{y}=\frac{A+B}{1-AB}$

Now,

$\frac{\tan 35^\circ+\tan 25^\circ}{1-\tan 35^\circ\tan 25^\circ}=\tan(35^\circ+25^\circ)$

$=\tan 60^\circ$

$=\sqrt{3}$

Therefore,

$\frac{x}{y}=\sqrt{3}$

So,

$x=\sqrt{3}y$

Since $\sqrt{3}>1$, according to the given option pattern, we get

$x>y$