Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$ 

Question: Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$



My attempt:
$$\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$$
$$ \frac{\frac {1}{\cos(x)} - \frac{1}{\sin(x)}}{\frac{\sin(x)}{\cos(x)} - \frac{\cos(x)}{\sin(x)}} $$
$$ \frac{\sin(x)-\cos(x)}{\sin^2(x)-\cos^2(x)}$$
$$ \frac{(\sin(x)-\cos(x))}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))} $$
$$ \frac{1}{\sin(x)+\cos(x)} $$
Now this is where I am stuck , I thought of multiplying the numerator and denominator by $$ \frac{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}}{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}} $$ but that did not work out well..

Start by multiplying both sides by the denominators, so that you get
$$\sec(x)^2-\csc(x)^2  = \tan(x)^2-\cot(x)^2.$$
Now starting from the left-hand side :
\begin{align*}\sec(x)^2-\csc(x)^2 & =\frac{1}{\cos(x)^2}-\frac{1}{\sin(x)^2} \\ & = \frac{1}{\cos(x)^2}-1-\frac{1}{\sin(x)^2}+1 \\ & = \frac{1-\cos(x)^2}{\cos(x)^2}-\frac{1-\sin(x)^2}{\sin(x)^2} \\ & = \frac{\sin(x)^2}{\cos(x)^2}-\frac{\cos(x)^2}{\sin(x)^2}.\end{align*}
