<s>[INST] 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 $1-\dfrac{\sin^2 x}{1+\frac{\cos x}{\sin x}}$
 [/INST] $$1-\frac{\sin^2 x}{1+\cot x}-\frac{\cos^2 x}{1+\tan x}=\sin x \cos x.$$
$$1-\frac{\sin^2 x}{1+\frac{\cos x}{\sin x}}-\frac{\cos^2 x}{1+\frac{\sin x}{\cos x}}=\sin x \cos x.$$
$$1-\frac{\sin^3 x}{\cos x+ \sin x}-\frac{\cos^3 x}{\sin x+\cos x}=\sin x \cos x.$$
$$\frac{(\sin x+\cos x)-(\sin^3 x+\cos^3 x)}{\sin x+\cos x}=\sin x\cos x$$
$$\frac{(\sin x+\cos x)-(\sin x+\cos x)(\sin^2 x-\sin x\cos x+\cos^2 x)}{\sin x+\cos x}=\sin x\cos x$$
$$\frac{(\sin x+\cos x)-(\sin x+\cos x)(1-\sin x\cos x)}{\sin x+\cos x}=\sin x\cos x$$
$$\frac{(\sin x+\cos x)(1-(1-\sin x\cos x))}{\sin x+\cos x}=\sin x\cos x$$
$$1-(1-\sin x\cos x)=\sin x\cos x$$
$$1-1+\sin x\cos x=\sin x\cos x$$
$$\sin x\cos x=\sin x\cos x$$
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