<s>[INST] 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.
 [/INST] Suppose $\sin x \neq 0$. Then we have that $\csc x = \dfrac{1}{\sin x}$ and $\dfrac{\cos x}{\sin x} = \cot x$ 
This implies in $\cos x = \dfrac{a-b}{a+b}$. 
Therefore, $\csc x \cdot \cot x = \dfrac {\cos x}{\sin^2 x} = \dfrac {\cos x}{1-\cos^2 x}.$
Notice that $1- \cos^2 x = 1 - \left(\dfrac{a-b}{a+b}\right)^2 = \dfrac{4ab}{(a+b)^2}$.
Then, $\dfrac {\cos x}{1-\cos^2 x}=\dfrac{(a-b)(a+b)}{4ab}=\dfrac{a^2-b^2}{4ab}.$
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