Proving trigonometric Identity: $\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x}$

$\frac{1+\sin x}{\cos x}$

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

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

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

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

$=\frac{1+\sin x+\cos x}{1-\sin x+\cos x}$

Using $\cos^2x+\sin^2x=1$, we have $$ (1+\sin x)(1-\sin x+\cos x)=1-\sin^2x+(1+\sin x)\cos x=\cos x(\cos x+1+\sin x).$$ Now divide the LHS by $1-\sin x+\cos x$ and the RHS by $\cos x$ to get the result.