<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] $$
\frac{a+b}{\csc x}=\frac{a-b}{\cot x}=k \qquad\implies\qquad
\begin{align}
a+b&=k\csc x \\
a-b&=k\cot x
\end{align}$$
Therefore,
$$\begin{align}
a^2 - b^2 &= ( a+ b )( a - b ) &&= k^2 \cdot \csc x \cot x \\
4 a b &= ( a + b )^2 - ( a - b )^2 = k^2 (\csc^2x-\cot^2x)&&=k^2\cdot 1
\end{align}$$
The result follows. $\square$
</s>