Prove the following identities (1-16)
Question:

Prove the following identities (1-16)

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

Solution:

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

$\mathrm{LHS}=1-\frac{\sin ^{3} x}{\sin x+\cos x}-\frac{\cos ^{3} x}{\sin x+\cos x}$

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

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

$=\left(1-\sin ^{2} x-\cos ^{2} x+\sin x \cos x\right)$

$=(1-1+\sin x \cos x)$

$=\sin x \cos x$

= RHS

Hence proved.