Proving the identity $\frac{1}{1-\cos3x} + \frac{1}{1+\cos3x}=2\csc^2(3x)$ [closed]

Question

I am trying to figure out the trig identity to prove the following:

$$\frac{1}{1-\cos3x} + \frac{1}{1+\cos3x} = 2\csc^2(3x)$$

I just cannot seem to figure it out. Thank you for your help.

Answer 1

Hint: Try to get the common denominator, and use the fact that $\sin^2x+\cos^2x=1$.

Answer 2

\begin{align*} \frac{1}{1-\cos(3x)}+\frac{1}{1+\cos(3x)}&=\frac{(1+\cos(3x))+(1-\cos(3x))}{1-\cos^2(3x)}\\ &=\frac{2}{1-\cos^2(3x)}\\ &=\frac{2}{\sin^2(3x)}\\ &=2\csc^2(3x) \end{align*}