<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] Rearrange and Cross multiply 
it is required to prove that
$$\dfrac{\cos x +1 +\sin x}{1-\sin x +\cos x} = \dfrac{1+\sin x}{\cos x}$$ or
$$ \cos^2 x + \cos x (1 +\sin x ) = (1-\sin^2 x) + \cos x (1 +\sin x ) $$ or
$$  \cos x (1 +\sin x ) =  \cos x (1 +\sin x ). $$
</s>