Question

Prove the following identities (1-16)

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

Answer 1

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


$$\begin{gathered} \mathrm{LHS}=1-\frac{{\sin }^{3}x}{\sin x+\cos x}-\frac{{\cos }^{3}x}{\sin x+\cos x} \\ =\frac{\sin x+\cos x-\left({\sin }^{3}x+{\cos }^{3}x\right)}{\sin x+\cos x} \\ =\frac{\left(\sin x+\cos x\right)\left(1-{\sin }^{2}x-{\cos }^{2}x+\sin x\cos x\right)}{\sin x+\cos x} \\ =\left(1-{\sin }^{2}x-{\cos }^{2}x+\sin x\cos x\right) \\ =\left(1-1+\sin x\cos x\right) \\ =\sin x\cos x \end{gathered}$$

= RHS
Hence proved.