Answer
$$\frac{1-\cos x}{1+\cos x}=(\cot x-\csc x)^2$$
The trigonometric expression is an identity.
Work Step by Step
$$\frac{1-\cos x}{1+\cos x}=(\cot x-\csc x)^2$$
The right side comprises of $\cot x$ and $\csc x$ so we need to deal with it first.
$$A=(\cot x-\csc x)^2$$
$$A=\cot^2x-2\cot x\csc x+\csc^2x$$
$$A=(\cot^2x+\csc^2x)-2\cot x\csc x$$
- For $\cot^2x+\csc^2x$, we use the Pythagoren Identity:
$$\csc^2x=1+\cot^2x$$
Therefore, $$\cot^2x+\csc^2x=\cot^2x+1+\cot^2x=2\cot^2x+1$$
Now we transform $\cot x$: $$\cot x=\frac{\cos x}{\sin x}$$
So, $$\cot^2x+\csc^2x=2\times\frac{\cos^2x}{\sin^2x}+1=\frac{2\cos^2x+\sin^2x}{\sin^2x}\hspace{1cm}(1)$$
- For $2\cot x\csc x$, we transform both $\cot x$ and $\csc x$:
$$\cot x=\frac{\cos x}{\sin x}$$
$$\csc x=\frac{1}{\sin x}$$
Therefore, $$2\cot x\csc x=2\times\frac{\cos x}{\sin x}\times\frac{1}{\sin x}=\frac{2\cos x}{\sin^2 x}\hspace{1cm}(2)$$
Combine $(1)$ and $(2)$ into $A$:
$$A=\frac{2\cos^2x+\sin^2x}{\sin^2x}-\frac{2\cos x}{\sin^2 x}$$
$$A=\frac{2\cos^2x+\sin^2x-2\cos x}{\sin^2x}$$
$$A=\frac{(\cos^2x+\sin^2x)+\cos^2x-2\cos x}{\sin^2 x}$$
$$A=\frac{\cos^2x-2\cos x+1}{\sin^2 x}$$ (for $\cos^2x+\sin^2x=1$)
$$A=\frac{(\cos x-1)^2}{\sin^2 x}$$
We find the left side having only $\cos x$, so we need to change the denominator also to $\cos x$ by $\sin^2x=1-\cos^2x$
$$A=\frac{(\cos x-1)^2}{1-\cos^2x}$$
$$A=\frac{(1-\cos x)^2}{(1-\cos x)(1+\cos x)}$$ (for $(a-b)^2=(b-a)^2$ and $a^2-b^2=(a-b)(a+b)$)
$$A=\frac{1-\cos x}{1+\cos x}$$
The trigonometric expression is thus an identity.
