how to show $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ I know that $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ is correct. But having some hard time proofing it using trig relations. Some of the relations I used are 
$$
\sin x \cos x= \frac{1}{2} \sin(2 x)\\
\sin^2 x + \cos^2 x= 1\\
\sin^2 x  = \frac{1}{2} - \frac{1}{2} \cos(2 x)\\
\cos(2 x) = 2 \cos^2(x)-1
$$
And others. I tried starting with multiplying numerator and denominator with either $\cos x$ or $\sin x$ and try to simplify things, but I seem to be going in circles. 
Any hints how to proceed?

You have to prove that
$$
\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}.\tag{1}
$$
Note that $(1)$ may be expressed as follows by cross-multiplying:
$$
\underbrace{(\cos x)(1+\cos x+\sin x)}_{\text{LHS}}=\underbrace{(1+\sin x)(1+\cos x-\sin x)}_{\text{RHS}}.\tag{2}
$$
Now all you have to do is expand the LHS and RHS to see that they are equivlant:
We have
\begin{align}
\text{LHS} 
&= (\cos x)(1+\cos x+\sin x)\tag{definition}\\[0.5em]
&= \cos x +\cos^2(x)+\cos x\sin x\tag{expand}
\end{align}
and
\begin{align}
\text{RHS} &= (1+\sin x)(1+\cos x-\sin x)\tag{definition}\\[0.5em]
&= 1+\cos x-\sin x+\sin x+\sin x\cos x-\sin^2(x)\tag{expand}\\[0.5em]
&= 1+\cos x+\sin x\cos x-\sin^2(x).\tag{simplify}
\end{align}
Now compare the LHS and RHS. We clearly have that
$$
\cos^2(x)=1-\sin^2(x)\Longleftrightarrow \cos^2(x)+\sin^2(x)=1,
$$
and we know this famous result to be true (or easily established otherwise). Hence, we have proved $(1)$.
