<s>[INST] 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?
 [/INST] $$\frac{1 + \cos x + \sin x}{1 + \cos x - \sin x}$$
$$= \frac{(1 + \cos x + \sin x)(1 + \cos x + \sin x)}{(1 + \cos x - \sin x)(1 + \cos x + \sin x)}$$
$$= \frac{1 + \cos^2 x + \sin^2 x + 2\cos x + 2\sin x + 2\sin x \cos x}{1 + 2\cos x + \cos^2 x - \sin^2 x}$$
$$= \frac{2(1 + \cos x + \sin x + \sin x \cos x)}{1 + 2\cos x + 2\cos^2 x - 1}$$
$$= \frac{2(1 + \cos x)(1 + \sin x)}{2\cos x(1 + \cos x)}$$
$$= \frac{1 + \sin x}{\cos x}$$
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