Is it generally true that $\arcsin \theta + \arccos \theta = \frac{\pi}{2}$? 
Verify that: $\arcsin \theta+\arccos \theta=\frac{\pi}{2}.$ (1)

How can one verify (1) when it is not generally true? We can rewrite (1) as 

Verify that if $\sin u = \cos v$, then $u+v=\frac{\pi}{2}$.

What about $\sin \frac{2\pi}{3}$ and $\cos \frac{\pi}{6}$? $\sin \frac{2\pi}{3} = \cos \frac{\pi}{6}$, but $\frac{2\pi}{3}+\frac{\pi}{6} \neq \frac{\pi}{2}$.We may try to verify (1) as:
Let $\arcsin \theta=u$ and $\arccos \theta=v$, then $\sin u=\cos v = \theta.$ We can write 
$$\sin(u+v)=\sin u \cos v+ \cos u \sin v,$$
which becomes 
$$\sin(u+v)=\theta^2+ \cos u \sin v.$$
Now, $\cos u=\pm \sqrt{1-\theta^2}$, and $\sin v=\pm \sqrt{1-\theta^2}$. For $\cos u= \sqrt{1-\theta^2}$ and $\sin v= \sqrt{1-\theta^2}$, we have
$$\sin(u+v)= \theta^2+(1-\theta^2)=1,$$
$$\therefore u+v=\arcsin \theta + \arccos \theta = \arcsin(1)=\frac{\pi}{2}.$$ But $\arcsin(1)$ is also equal to $\frac{\pi}{2}+2n\pi, n \in \mathbb{Z}.$ What about that? And what about the cases when $\cos u = -\sqrt{1-\theta^2}$ and $\sin u = \sqrt{1-\theta^2}$ and the other way? Then $\sin(u+v)=2\theta^2-1=1\ \mbox{iff}\ \theta =\pm 1.$ 
I don't know where I am making a mistake, what assumptions are wrong, or where my reasoning goes wrong. Please help.

The error is in thinking that $\arcsin$ and $\arccos$ are multivalued functions that give all possible values that will yield the given sine or cosine. Instead the range of $\arcsin$ is $[-\pi/2,\pi/2]$ and the range of $\arccos$ is $[0,\pi]$, and the functions take no values outside these ranges.
