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 and divide $ \frac{1}{\sin{x}+\cos{x}}$ by $ \frac{1}{\sin{x}\cos{x}}$ then in the numerator substitute $1$ by $sin^2{x} + cos^2{x}$
