Prove an trigonometric identity. Can someone help me by solving it? $$1-\frac{\sin^2 x}{1+\cot x}-\frac{\cos^2 x}{1+\tan x}=\sin x \cos x.$$
I cannot come to the result. I make that like $1-\dfrac{\sin^2 x}{1+\frac{\cos x}{\sin x}}$

We can transform the left side to the right.
$$\underbrace{\color{white}{\underline{\color{black}{1-\dfrac{\sin^2x}{1+\cot x}-\dfrac{\cos^2 x}{1+\tan x}}}}}_{\displaystyle\cal I}=\sin x\cos x. \\ \eqalign{
{\cal I} \ : & \; \  1-\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}-\dfrac{\cos^2x}{\dfrac{\cos x+\sin x}{\cos x}} \\
&= \dfrac{\sin x+\cos x-\sin^3 x-\cos^3x}{\cos x+\sin x} \\
&= \dfrac{\sin x+\cos x-(\sin x+\cos x)\left(\sin^2x-\sin x\cos x+\cos^2x      \right)}{\cos x+\sin x} \\
&=\require{cancel}\dfrac{(\cancel{\cos x+\sin x })\left(1-\sin^2x-\cos^2x+\sin x\cos x\right)}{\cancel{\cos x+\sin x }}\\
&=1-1+\sin x\cos x=\color{lightblue}{\underline{\color{black}{\sin x\cos x}}}
\;\checkmark}$$
