<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] $$
\begin{align}
1-\frac{\sin^2(x)}{1+\cot(x)}-\frac{\cos^2(x)}{1+\tan(x)}
&=\frac{(1+\tan(x))-\sin^2(x)\tan(x)-\cos^2(x)}{1+\tan(x)}\tag{1}\\
&=\frac{1+\cos^2(x)\tan(x)-\cos^2(x)}{1+\tan(x)}\tag{2}\\
&=\frac{\cos^2(x)\tan(x)+\sin^2(x)}{1+\tan(x)}\tag{3}\\
&=\frac{\cos^2(x)\tan(x)+\cos^2(x)\tan^2(x)}{1+\tan(x)}\tag{4}\\[4pt]
&=\cos^2(x)\tan(x)\tag{5}\\[12pt]
&=\sin(x)\cos(x)\tag{6}
\end{align}
$$
Explanation:
$(1)$: put everything over a common denominator
$(2)$: $\cos^2(x)=1-\sin^2(x)$
$(3)$: $\sin^2(x)=1-\cos^2(x)$
$(4)$: $\sin^2(x)=\cos^2(x)\tan^2(x)$
$(5)$: factor out $1+\tan(x)$ from the numerator and cancel
$(6)$: $\sin(x)=\cos(x)\tan(x)$
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