TITLE: A trigonometry equation problem
QUESTION [5 upvotes]: Let $a,b \in \left(0,\frac{\pi}{2}\right)$, satisfying
$$
\frac{1-\cos{2a}}{1+\sin{a}}+\frac{1+\cos{2a}}{1+\cos{b}}=\frac{1-\cos{(2a+2b)}}{1+\sin{(a+2b)}}
$$
Prove that:
$$
a+b=\frac{\pi}{2}
$$

REPLY [4 votes]: Let, without loss of generality, $a=\frac{\pi}{2}-(b-c)$, 
then by using the fact that you can suck up $\frac{\pi}{2}$ (and thereby change $\sin$ to $\cos$) as well as $\pi$ (and thereby change a sign of the trigonometric function) you can get rid of all $\pi$'s and $a$'s:
$$
\frac{1-\cos{2a}}{1+\sin{a}}+\frac{1+\cos{2a}}{1+\cos{b}}=\frac{1-\cos{(2a+2b)}}{1+\sin{(a+2b)}}
$$
$$\Longrightarrow$$
$$
\frac{1+\cos{(2(b-c))}}{1+\cos{(b-c)}}+\frac{1-\cos{(2(b-c))}}{1+\cos{b}}=\frac{1+\cos{(2c)}}{1+\cos{(b+c)}}
$$
Now it is acutally clear that $a+b=\frac{\pi}{2}$ is a solution, because if you plug in $c=0$, all three denomiators become $1+\cos{(b)}$ and the equation reads $2=2$.
If you want to go futher you can use $$\cos ( b \pm c ) = \cos (b)  \cos (c) \mp \sin (b)  \sin (c)$$ to isolate the two trigonometric functions of $c$, substitue $t=\tan{\frac{c}{2}}$ to get polynomial expression in $t$, put everything on one equal denominator and solve the thing. This will amount to $t=\tan{(0)}=0$.