Question with Solution

On Putting $x=\tan A,y=\tan B$

we get $\sec A+\sec B=\lambda (\tan A\sec B-\tan B\sec A)$

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

$\tan \left(\frac{A-B}{2}\right)=\frac{1}{\lambda }$

${\tan }^{-1}x-{\tan }^{-1}y=2{\tan }^{-1}\frac{1}{\lambda }$

On differentiating $\frac{1}{1+x^2}-\frac{1}{1+y^2}\frac{dy}{dx}=0$