Answer
$$\frac{1-\cos\theta}{1+\cos\theta}=2\csc^2\theta-2\csc\theta\cot\theta-1$$
The expression is proved below to be an identity.
Work Step by Step
$$\frac{1-\cos\theta}{1+\cos\theta}=2\csc^2\theta-2\csc\theta\cot\theta-1$$
We actually figure out from exercise 73 that
$$\frac{1-\cos x}{1+\cos x}=\csc^2x-2\csc x\cot x+\cot^2x$$
Applying here, we have the identity
$$\frac{1-\cos\theta}{1+\cos\theta}=\csc^2\theta-2\csc\theta\cot\theta+\cot^2\theta$$
That means to prove the requirement of the exercise, we only need to prove
$$2\csc^2\theta-2\csc\theta\cot\theta-1=\csc^2\theta-2\csc\theta\cot\theta+\cot^2\theta\hspace{1cm}(1)$$
Thus, we would start with the left side.
$$A=2\csc^2\theta-2\csc\theta\cot\theta-1$$
$$A=\csc^2\theta-2\csc\theta\cot\theta+(\csc^2\theta-1)$$
- From Pythagorean Identity: $\csc^2\theta=\cot^2\theta+1$
Replace this into the second $\csc^2\theta$ of $A$:
$$A=\csc^2\theta-2\csc\theta\cot\theta+(\cot^2\theta+1-1)$$
$$A=\csc^2\theta-2\csc\theta\cot\theta+\cot^2\theta$$
Therefore, $(1)$ is proved to be an identity. That means $$\frac{1-\cos\theta}{1+\cos\theta}=2\csc^2\theta-2\csc\theta\cot\theta-1$$
is also an identity.
