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..
 A: HINT:
$$\sec^2x-\tan^2x=1=\csc^2x-\cot^2x$$
$$\iff\sec^2x-\csc^2x=\tan^2x-\cot^2x$$
$$\sec^2x-\csc^2x=(\sec x-\csc x)(\sec x+\csc x)$$
Can you take it from here?
A: Take the right hand side:
$$\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}=\frac{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}}{\frac{1}{\cos(x)} + \frac{1}{\sin(x)}} = \frac{\sin^2x+\cos^2x}{\sin x+\cos x}=\frac{1}{\sin x+\cos x} $$
and this is equal to the expression you found for the left hand side.
A: multiply and divide $ \frac{1}{\sin{x}+\cos{x}}$ by $ \frac{1}{\sin{x}\cos{x}}$ then in the numerator substitute $1$ by $sin^2{x} + cos^2{x}$
A: 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*}
A: 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.
A: Cross  multiplying we need to prove that:
$$\sec ^2 x -\csc ^2 x  = \tan  ^2 x -\cot  ^2 x $$
or to prove that
$$1+ \tan ^2 x -(1 +\cot ^2 x ) = \tan  ^2 x -\cot  ^2 x  !! $$
