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.
 A: 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, $\dfrac {\cos x}{1-\cos^2 x}=\dfrac{(a-b)(a+b)}{4ab}=\dfrac{a^2-b^2}{4ab}.$
A: $$
\frac{a+b}{\csc x}=\frac{a-b}{\cot x}=k \qquad\implies\qquad
\begin{align}
a+b&=k\csc x \\
a-b&=k\cot x
\end{align}$$
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. $\square$
A: 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 $\frac{a+b}{a-b}=\frac{\csc x}{\cot x}$,
$$\frac{a^2-b^2}{4ab}=\frac1{\frac{\csc x}{\cot x}-\frac{\cot x}{\csc x}}
=\frac{\csc x\cot x}{\csc^2 x-\cot^2x}=\csc x\cot x$$
A: 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.
A: Like the formulas mentioned here in https://www.onlinemath4all.com/properties-of-proportion.html,
$$\dfrac{\csc x}{a+b}=\dfrac{\cot x}{a-b}=p\sqrt{\dfrac{\csc^2x-\cot^2x}{(a+b)^2-(a-b)^2}}$$
where $p=\pm1\implies p^2=1$
