# If tan(pi/2sintheta) = cot(pi/2costheta) then sum of all possible values of sintheta + costheta equal to?

$0$
If $\setminus \tan A = \tan B$, then $A - B = n \setminus \pi$, where $n \setminus \in \setminus m a t h \boldsymbol{Z}$. Since you can write the given equation as $\setminus \tan \left(\setminus \frac{\setminus \pi}{2} \setminus \sin \setminus \theta\right) = \setminus \tan \left(\setminus \frac{\setminus \pi}{2} - \setminus \frac{\setminus \pi}{2} \setminus \cos \setminus \theta\right)$, we have that $\setminus \frac{\setminus \pi}{2} \setminus \sin \setminus \theta - \setminus \frac{\pi}{2} + \setminus \frac{\pi}{2} \setminus \cos \setminus \theta = n \setminus \pi$, or $\setminus \sin \setminus \theta + \setminus \cos \setminus \theta = \frac{1}{2} + n$, where $n \setminus \in \setminus m a t h \boldsymbol{Z}$.
Of course, $\setminus \sin \setminus \theta + \setminus \cos \setminus \theta$ can only lie between $- \setminus \sqrt{2}$ and $\setminus \sqrt{2}$, so the only possible values are at $n = 0$ and $n = - 1$, and they are $\frac{1}{2}$ and $- \frac{1}{2}$. That's it!