Question

The degree of the differential equation satisfying the relation $\sqrt{1+x^2}+\sqrt{1+y^2}=\lambda (x\sqrt{1+y^2}-y\sqrt{1+x^2})$ is

• A. 1
• B. 2
• C. 3
• D. none of these

Answer 1

The correct option is A 1

On Putting $x=\tan A,y=\tan B$ we get
$\sec A+\sec B=\lambda (\tan A\sec B-\tan B\sec A)\cos A+\cos B=\lambda (\sin A-\sin B)2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=\lambda \times 2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)\tan \left(\frac{A-B}{2}\right)=\frac{1}{\lambda }{\tan }^{-1}x-{\tan }^{-1}y=2{\tan }^{-1}\frac{1}{\lambda }$
On differentiating $\frac{1}{1+x^2}-\frac{1}{1+y^2}\frac{dy}{dx}$ = 0