Question

If $y=\frac{\cos x-\sin x}{\cos x+\sin x}$, show that $\frac{dy}{dx}+y^2+1=0$.

Answer 1

Given that,

$y=\frac{\cos x-\sin x}{\cos x+\sin x}$

on rationlize

$y=\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\times \left(\frac{\cos x-\sin x}{\cos x-\sin x}\right)$

$y=\frac{(\cos x-\sin x{)}^2}{{\cos }^{2}x-{\sin }^{2}x}$

$y=\frac{{\cos }^{2}x+{\sin }^{2}x-2\sin x\cos x}{\cos 2x}$

$y=\frac{1-\sin 2x}{\cos 2x}$

$y=\frac{1}{\cos 2x}-\frac{\sin 2x}{\cos 2x}$

$y=\sec 2x-\tan 2x$

on diff. and we get

$\frac{dy}{dx}=\frac{d}{dx}(\sec 2x-\tan 2x)$

$=\sec 2x\tan 2x\times 2-{\sec }^{2}2x\times 2$

$=2\sec 2x\tan 2x-2{\sec }^{2}2x$

now, L.H.S.

$\frac{dy}{dx}+y^2+1$

$2\sec 2x\tan 2x-2{\sec }^{2}2x+(\sec 2x-\tan 2x{)}^2+1$

$2\sec 2x\tan 2x-2{\sec }^{2}2x+{\sec }^{2}2x+{\tan }^{2}2x-2\sec 2x\tan 2x+1$

$-{\sec }^{2}2x+{\tan }^{2}2x+1$

$-({\sec }^2+{\tan }^{2}2x)+1$

$\because {\sec }^{2}A-{\tan }^{2}A=1$

$-1+1=0$

R.H.S.

Hence proved.