A trigonometry equation problem 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:
$$
a+b=\frac{\pi}{2}
$$
 A: The following calculations, show the non-uniqueness of the statement to be proven and therefore that the proposition is not true, when we forget the restriction $a,b \in \left(0,\frac{\pi}{2}\right)$ (Thanks to Christian). Then $a+b=\frac\pi2$ is a sufficient, but not a necessary condition for $f(a,b) = 0$, because of the existence of other zeros. $f(a,b)$ can be splitted in at least $3$ factors and so $f(a,b)=0$ implies $\frac12\left(a+b-\frac\pi 2\right)=\pi n$ and/or $\frac12(a-\frac\pi2)=\pi m$, plus some more weird solutions, where a nice general expression is currently lacking (or might not exist), although some particular values can be specified. Here it is:
When you plot 
$$
\begin{eqnarray}
f(a,b)&=&(1+\cos(b))(1+\sin(a+2b))(1-\cos 2a)\\
&+&(1+\sin a)(1+\sin(a+2b))(1+\cos2a)\\
&-&(1+\cos(b))(1+\sin a)(1-\cos(2a+2b))=0
\end{eqnarray}
$$
you'll clearly see the $a+b=\frac{\pi}{2}$-line, among other possible solutions (some of them are briefly discussed below in the EDIT):
$\hskip1.7in$
W|A helped to expand it
to 
$$
\sin(\frac a2-\frac\pi4) \color{red}{\sin(\frac a2+\frac b2-\frac\pi 4)} 
\biggr\{
\sin(a-\frac52 b)+3 \sin(a-\frac b2)+3 \sin(a+\frac b2)\\+6 \sin(a+\frac32 b)-\sin(3 a+\frac32 b)+2 \sin(a+\frac52 b)-2 \cos(2 a-\frac b2)\\ -\cos(2 a+\frac b2)-3 \cos(2 a+\frac32 b)+7 \cos(\frac b2)+2 \cos(\frac32 b)+\cos(\frac52 b)
\biggr\}=0 \tag{*}
$$
and then it is obvious, that $\frac12\left(a+b-\frac\pi 2\right)=\pi n$ solves this equation, as well as $\frac12(a-\pi/2)=\pi m$ does. Other solutions are visible, but not obvious.
EDIT Wolfram kindly provides some more
particular roots of the $\{\cdots\}$-part of $(*)$ like
$$
a=2\pi c_1+\pi \; \;, b=-4\left(\pi c_2+\tan^{-1}(x_k)\right),
$$
where 


*

*$x_{0,1}=1\pm\sqrt{2}\;$ (roots of $x^2-2x-1\;$) or 

*$x_{2,3,4,5}$, one of $4$ real roots of $x^8-8 x^7-20 x^6-8 x^5+22 x^4+8x^3-20 x^2+8 x+1$. 



The converse of the statement is much easier to proof:
Substitute $b=\pi/2-a$ to get:
$$
\begin{eqnarray}
\frac{1-\cos{2a}}{1+\sin{a}}+\frac{1+\cos{2a}}{1+\cos({\frac\pi2-a})}
&=&\frac{1-\cos{(2a+\pi-2a)}}{1+\sin{(a+\pi-2a)}}\\
&=&\frac{1-\cos{(\pi)}}{1+\sin{(\pi-a)}}\\
\frac{1-\cos{2a}}{1+\sin{a}}+\frac{1+\cos{2a}}{1+\sin a}&=&\frac{2}{1+\sin{a}}\\
\frac{2}{1+\sin{a}}&=&\frac{2}{1+\sin{a}}\\
\end{eqnarray}
$$
A: 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 $t=\tan{(0)}=0$.
