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?

For $b\not=0$ we have:
$$\begin{align*}\frac{1+a}{b} &= \frac{1+a+b}{1-a+b}\qquad&\iff \\
(1+a)(1-a+b) &= b(1+a+b)\qquad&\iff\\
1-a^2+b(a+1) &= b^2 + b(a+1)\qquad&\iff\\
a^2+b^2&=1
\end{align*}$$
So your equation is an alternate way to characterize $\cos^2 x+\sin^2 x = 1$
