If $\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$, show that $\csc x \cot x=\frac{a^2-b^2}{4ab}$ 
If $$\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$$ 
  show that 
  $$\csc x \cot x=\frac{a^2-b^2}{4ab}$$

(original problem image)
I tried getting rid of the denominator and then expanding the given equation, but couldn't get to the answer.

Rearrange
$$\frac{a^2-b^2}{4ab}=\frac{(a+b)(a-b)}{(a+b)^2-(a-b)^2}
=\frac{1}{\frac{a+b}{a-b}-\frac{a-b}{a+b}}$$
Then, substitute $\frac{a+b}{a-b}=\frac{\csc x}{\cot x}$,
$$\frac{a^2-b^2}{4ab}=\frac1{\frac{\csc x}{\cot x}-\frac{\cot x}{\csc x}}
=\frac{\csc x\cot x}{\csc^2 x-\cot^2x}=\csc x\cot x$$
