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 $1-\dfrac{\sin^2 x}{1+\frac{\cos x}{\sin x}}$
 A: A starting point:
$$
\frac{1}{1+\cot x}=\frac{\sin x}{\sin x+\cos x},\qquad
\frac{1}{1+\tan x}=\frac{\cos x}{\sin x+\cos x}
$$
Then
$$
\sin^3x+\cos^3x=(\sin x+\cos x)(\sin^2x -\sin x\cos x+\cos^2x)
=(\sin x+\cos x)(1-\sin x\cos x)
$$
A: We can transform the left side to the right.
$$\underbrace{\color{white}{\underline{\color{black}{1-\dfrac{\sin^2x}{1+\cot x}-\dfrac{\cos^2 x}{1+\tan x}}}}}_{\displaystyle\cal I}=\sin x\cos x. \\ \eqalign{
{\cal I} \ : & \; \  1-\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}-\dfrac{\cos^2x}{\dfrac{\cos x+\sin x}{\cos x}} \\
&= \dfrac{\sin x+\cos x-\sin^3 x-\cos^3x}{\cos x+\sin x} \\
&= \dfrac{\sin x+\cos x-(\sin x+\cos x)\left(\sin^2x-\sin x\cos x+\cos^2x      \right)}{\cos x+\sin x} \\
&=\require{cancel}\dfrac{(\cancel{\cos x+\sin x })\left(1-\sin^2x-\cos^2x+\sin x\cos x\right)}{\cancel{\cos x+\sin x }}\\
&=1-1+\sin x\cos x=\color{lightblue}{\underline{\color{black}{\sin x\cos x}}}
\;\checkmark}$$
A: $$
\begin{align}
1-\frac{\sin^2(x)}{1+\cot(x)}-\frac{\cos^2(x)}{1+\tan(x)}
&=\frac{(1+\tan(x))-\sin^2(x)\tan(x)-\cos^2(x)}{1+\tan(x)}\tag{1}\\
&=\frac{1+\cos^2(x)\tan(x)-\cos^2(x)}{1+\tan(x)}\tag{2}\\
&=\frac{\cos^2(x)\tan(x)+\sin^2(x)}{1+\tan(x)}\tag{3}\\
&=\frac{\cos^2(x)\tan(x)+\cos^2(x)\tan^2(x)}{1+\tan(x)}\tag{4}\\[4pt]
&=\cos^2(x)\tan(x)\tag{5}\\[12pt]
&=\sin(x)\cos(x)\tag{6}
\end{align}
$$
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
$(6)$: $\sin(x)=\cos(x)\tan(x)$
A: $$1-\frac{\sin^2 x}{1+\cot x}-\frac{\cos^2 x}{1+\tan x}=\sin x \cos x.$$
$$1-\frac{\sin^2 x}{1+\frac{\cos x}{\sin x}}-\frac{\cos^2 x}{1+\frac{\sin x}{\cos x}}=\sin x \cos x.$$
$$1-\frac{\sin^3 x}{\cos x+ \sin x}-\frac{\cos^3 x}{\sin x+\cos x}=\sin x \cos x.$$
$$\frac{(\sin x+\cos x)-(\sin^3 x+\cos^3 x)}{\sin x+\cos x}=\sin x\cos x$$
$$\frac{(\sin x+\cos x)-(\sin x+\cos x)(\sin^2 x-\sin x\cos x+\cos^2 x)}{\sin x+\cos x}=\sin x\cos x$$
$$\frac{(\sin x+\cos x)-(\sin x+\cos x)(1-\sin x\cos x)}{\sin x+\cos x}=\sin x\cos x$$
$$\frac{(\sin x+\cos x)(1-(1-\sin x\cos x))}{\sin x+\cos x}=\sin x\cos x$$
$$1-(1-\sin x\cos x)=\sin x\cos x$$
$$1-1+\sin x\cos x=\sin x\cos x$$
$$\sin x\cos x=\sin x\cos x$$
A: Your question
$$\Rightarrow1-\frac{\sin^3x+\cos^3x}{\sin x+\cos x}=\sin x\cos x$$
$$\Rightarrow1-\frac{(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)}{\sin x+\cos x}=\sin x\cos x$$
$$\Rightarrow1-1+\sin x\cos x=\sin x\cos x\quad (\sin x+\cos x\ne0)$$
$$\Rightarrow\sin x\cos x=\sin x\cos x\quad (\sin x+\cos x\ne0)$$
which means the answers is this set of $x$:
$$\{x|x\in R,\sin x+\cos x\ne0\}$$
that is
$$\{x|x\in R,x\ne\frac{k\pi}{2},k\in Z\}$$
