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.

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


Use $\cos(2x)=\cos^2(x)-\sin^2(x)$ and $\sin(2x)=2\sin(x)\cos(x)$ to get $$\frac{\cos(2x)}{1+\sin(2x)}=\frac{\cos^2(x)-\sin^2(x)}{1+2\sin(x)\cos(x)}$$
Multiply top and bottom by $\frac{1}{\sin^2(x)}$ to get $$\frac{\cot^2(x)-1}{\frac{1}{sin^2(x)}+2\cot(x)}$$
Using $\sin^2(x)+\cos^2(x)=1$ we can get $$\frac{\cot^2(x)-1}{\frac{\sin^2(x)+\cos^2(x)}{sin^2(x)}+2\cot(x)}\ \ \ \ \ \ \ or\ \ \ \ \ \ \ \frac{\cot^2(x)-1}{1+\cot^2(x)+2\cot(x)}$$
By using $a^2-b^2=(a-b)(a+b)$ for the numerator and factoring the denominator, we get $$\frac{(\cot(x)-1)(\cot(x)+1)}{(\cot(x)+1)^2}$$ which equals to $$\frac{\cot(x)-1}{\cot(x)+1}$$
