Question

If $y=\frac{tanx+cotx}{tanx-cotx}$

Find $\frac{dy}{dx}$

Answer 1

$y=\frac{tanx+cotx}{tanx-cotx}=\frac{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}{\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}}=\frac{\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}}{\frac{{\sin }^{2}x-{\cos }^{2}x}{\sin x\cos x}}=\frac{1}{-\cos 2x}=-\sec 2x$

$\frac{dy}{dx}=-sec2x\tan 2x.2=-2sec2x\tan 2x$