<s>[INST] Proving this trig identity:$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$ 
$$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$$

What I've tried,
$$\frac{((1+\cos\theta)+(\sin\theta))((1+\cos\theta)+(\sin\theta))}{(1+\cos\theta-\sin\theta) (1+\cos\theta+\sin\theta)}$$
$$=\frac{(1+\cos\theta+\sin\theta)^2}{(1+\cos\theta)^2-\sin^2\theta}$$
After simplifying,
$$=\frac{2(1+\sin\theta \cos\theta + \sin\theta + \cos\theta)}{\cos\theta(\cos\theta+2)}$$
I cant carry on further. Thus, any kind assistance would be much appreciated.
 [/INST] Multiplying and dividing by $(1+\sin\theta -\cos \theta)$ yields
$$\frac{(1+\sin \theta)^2-\cos^2 \theta}{1-\cos^2\theta -\sin^2\theta +2\sin\theta \cos \theta}= \frac{1+\sin^2 \theta + 2\sin \theta -\cos^2 \theta}{2\sin \theta \cos \theta}$$
Substituting $-\cos^2 \theta = \sin^2 \theta -1$, we obtain
$$ \frac{1+\sin^2 \theta +2\sin \theta +\sin^2 \theta -1}{2 \sin \theta \cos \theta}= \frac{2 \sin \theta(1+\sin \theta)}{2 \sin \theta \cos \theta}$$
$$=\color{red} {\frac{1+\sin \theta}{\cos \theta}}$$
Edit: clarifications for OP.
Consider the initial expression multiplied and divided by $(1+\sin\theta -\cos \theta)$:
$$\frac{((1+\cos \theta+\sin \theta)(1+\sin \theta - \cos \theta)}{(1+\cos \theta - \sin \theta)(1+ \sin \theta - \cos \theta)}$$
Considering the terms grouped this way and using the fact that $(a+b)(a-b)=a^2-b^2$ we can conclude that
$$\frac{((1+\sin \theta)+\cos \theta)((1+\sin \theta) - \cos \theta)}{(1+(\cos \theta - \sin \theta))(1-(\cos \theta - \sin \theta)}=\frac{(1+\sin \theta)^2-\cos^2\theta}{1-(\cos \theta - \sin \theta)^2}$$
Expanding the square in the denominator leads to the LHS of the first equality.
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