Mathematics - Differential Equation Question with Solution | TestHub

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

$\begin{array}{rl}& \sec A+\sec B=\lambda (\tan A\sec B-\tan B\sec A)\\ & \cos A+\cos B=\lambda (\sin A-\sin B)\\ & \tan \left(\frac{A-B}{2}\right)=\frac{1}{\lambda }\\ & {\tan }^{-1}x-{\tan }^{-1}y=2{\tan }^{-1}\frac{1}{\lambda }\end{array}$

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