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..

Multiply the LHS by product of the reciprocal of RHS, and the RHS.
$$\begin{array}{lll}
\displaystyle\frac{\sec x - \csc x}{\tan x - \cot x}&=&\displaystyle\frac{\sec x - \csc x}{\tan x - \cot x}\cdot\frac{\sec x + \csc x}{\tan x + \cot x}\cdot\frac{\tan x + \cot x}{\sec x + \csc x}\\
&=&\displaystyle\frac{\sec^2 x - \csc^2 x}{\tan^2 x - \cot^2 x}\cdot\frac{\tan x + \cot x}{\sec x + \csc x}\\
&=&\displaystyle\frac{(\tan^2 x + 1) - (\cot^2 x+1)}{\tan^2 x - \cot^2 x}\cdot\frac{\tan x + \cot x}{\sec x + \csc x}\\
&=&\displaystyle\frac{\tan x + \cot x}{\sec x + \csc x}\\
\end{array}$$
The trick: if LHS=RHS then $LHS\times\frac{1}{RHS}$ will always equal 1, just as $\frac{1}{RHS}\times RHS$ will always equal 1.
