<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] Let us look at the RHS only: (It is fine to manipulate both sides but it is usually preferred if you manage to make one into the other.)
$$
\frac{\cot{\theta}-1}{\cot\theta +1} = \frac{\frac{\cos\theta}{\sin\theta}-1}{\frac{\cos\theta}{\sin\theta} +1}
$$
Now we multiply the numerator and denominator by $\sin\theta$ to get,
\begin{equation}
\frac{\cos\theta-\sin\theta}{\cos\theta +\sin\theta}.
\end{equation}
If we consider the double angle formulas for $\sin$ and $\cos$ i.e.
$$
\cos{2\theta} = \cos^2\theta-\sin^2\theta, \\ \sin 2\theta= 2\sin\theta\cos\theta,
$$
we can see that we can get a $\cos2\theta$ in the numerator of $\frac{\cos\theta-\sin\theta}{\cos\theta +\sin\theta}$ by multiplying by $\cos\theta+\sin\theta$ as the $\cos2\theta$ formula is a difference of two squares. Proceeding with this idea we get;
$$
\left(\frac{\cos\theta-\sin\theta}{\cos\theta +\sin\theta}\right)\left(\frac{\cos\theta+\sin\theta}{\cos\theta+\sin\theta}\right) = \frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta+2\sin\theta\cos\theta +\sin^2\theta}.
$$
Now the numerator is equal to $\cos2\theta$ so that is done and we only need to fix the denominator, by a quick inspection we see that we have a term $2\sin\theta\cos\theta$ which gives our $\sin2\theta$ term we desired and using the fact $\cos^2\theta+\sin^2\theta=1$ we find we get
$$
\frac{\cos2\theta}{1+\sin2\theta}
$$
which is what we wanted!
</s>