<s>[INST] Verifying $\frac{\cos{2\theta}}{1 + \sin{2\theta}} = \frac{\cot{\theta} - 1}{\cot{\theta} + 1}$ My math teacher gave us an equality involving trigonometric functions and told us to "verify" them. I tried making the two sides equal something simple such as "1 = 1" but kept getting stuck. I would highly appreciate if someone could show me (step by step) how to verify or solve this problem.

$$\frac{\cos{2\theta}}{1 + \sin{2\theta}} = \frac{\cot{\theta} - 1}{\cot{\theta} + 1}$$

 [/INST] Use
$\cos(2x)=\cos^2(x) - \sin^2(x)$ and 
$1+\sin(2x) = 1+2\sin(x)\cos(x) = \sin^2(x)+\cos^2(x)+2\sin(x)\cos(x)=(\sin(x)+\cos(x))^2$
So we have 
$$\frac{\cos(2x)}{1+\sin(2x)}=\frac{\cos^2(x) - \sin^2(x)}{(\sin(x)+\cos(x))^2}=\frac{\cos(x) - \sin(x)}{\cos(x) + \sin(x)}=\frac{\cot(x)-1}{\cot(x)+1}$$
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