Proving trigonometric Identity: $\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$ I would like to try and prove
$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$$
using $LHS=RHS$ methods, i.e. pick a side and rewrite it to make it identical to the other side.
I found a quick way by doing this:
$$LHS = \frac{1+\sin x}{\cos x} = \frac{1+\sin x}{\cos x} \cdot \frac{1 - \tan x + \sec x}{1 - \tan x + \sec x}= \frac{1+\sin x+\cos x}{1-\sin x+\cos x} = RHS$$
but I feel that this is not a good way because I am manipulating the denominator of the LHS somewhat artificially, because I know it must be, in the end, $1-\sin x+\cos x$.
Does anyone have a better way of doing this?

$\frac{1+\sin x}{\cos x}$
$=\frac{(1+\sin x)(1-\sin x+\cos x)}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{1-\sin^2 x+\cos x+\cos x\sin x}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{\cos^2 x+\cos x+\cos x\sin x}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{(\cos x)(1+\sin x+\cos x)}{(\cos x)(1-\sin x+\cos x)}$
$=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}$
