Question

If $y=\frac{{\sin }^{2}x}{1+\cot x}+\frac{{\cos }^{2}x}{1+\tan x}$ then $y^{{}^{\prime }}$ is equal to?

Answer 1

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

$y=\frac{{\sin }^{3}x}{\sin x+\cos x}+\frac{{\cos }^{3}x}{\sin x+\cos x}$

$\frac{(\sin x+\cos x{)}^3-3\sin x\cos x(\sin x+\cos x)}{\sin x+\cos x}$

$(\cos x+\sin x{)}^2-3\sin x\cos x$

$y=1-\sin x\cos x$

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

$y^{{}^{\prime }}=-\cos 2x$