<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] From $1+\cot^2 x= \csc^2 x$ we have 
\begin{eqnarray*}
(\color{red}{\csc x} +\cot x)(\csc x -\color{blue}{\cot x})  = 1
\end{eqnarray*}
Now sub 
\begin{eqnarray*}
\color{red}{\csc x} = \frac{a+b}{a-b} \cot x 
\end{eqnarray*}
and 
\begin{eqnarray*}
\color{blue}{\cot x} = \frac{a-b}{a+b} \csc x .
\end{eqnarray*}
This gives 
\begin{eqnarray*}
\cot x \left( \frac{a+b}{a-b} +1 \right)   \csc  x \left( 1-\frac{a-b}{a+b}\right) =1
\end{eqnarray*}
and the result follows with a bit of algebra.
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