Question

$\frac{d}{dx}\left(\frac{{\cot }^{2}x-1}{{\cot }^{2}x+1}\right)=$

Answer 1

Let $y=\frac{{\cot }^{2}x-1}{{\cot }^2+1}=\frac{{\cos }^{2}x-{\sin }^{2}x}{{\cos }^{2}x+{\sin }^{2}x}$

$=\cos 2x$ $(\because {\cos }^{2}x+{\sin }^{2}x=1)$

$\therefore y=\cos 2x$

$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{{\cot }^{2}x-1}{{\cot }^2+1}\right)=\frac{d}{dx}\cos 2x=-2\sin 2x$