Question

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

Answer 1

$$1-\frac{{\sin }^{2}x}{1+\cot x}-\frac{{\cos }^{2}x}{1+\tan x}=\sin x\cos x$$
$$\text{ LHS }=1-\frac{{\sin }^{3}x}{\sin x+\cos x}-\frac{{\cos }^{3}x}{\sin x+\cos x}$$
$$=\frac{\sin x+\cos x-({\sin }^{3}x+{\cos }^{3}x)}{\sin x+\cos x}$$
$$=\frac{(\sin x+\cos x)(1-{\sin }^{2}x-{\cos }^{2}x+\sin x\cos x)}{\sin x+\cos x}$$
$$=(1-{\sin }^{2}x-{\cos }^{2}x+\sin x\cos x)$$
$$=(1-1+\sin x\cos x)$$
$$=\sin x\cos x$$
= RHS
Hence proved.