Answer
$a.\quad 10$
$b.\displaystyle \quad-\frac{\pi}{4}$
Work Step by Step
$ a.\quad$
$\arctan 10=y \Leftrightarrow \tan y=10$ and $y\in(- \displaystyle \frac{\pi}{2},\frac{\pi}{2})$
so, $y=\arctan 10$ is such that $\tan y=\tan(\arctan 10)=10.$
$ b.\quad$
$a=\displaystyle \sin\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}$, and $\arcsin a=\sin^{-1}a$.
$\displaystyle \sin^{-1}(-\frac{\sqrt{2}}{2})=y \ \Leftrightarrow\ \displaystyle \sin y=-\frac{\sqrt{2}}{2}\ $ and $\ y\in[- \displaystyle \frac{\pi}{2},\frac{\pi}{2}]$
In $[- \displaystyle \frac{\pi}{2},\frac{\pi}{2}],\quad y=-\displaystyle \frac{\pi}{4}$ is such that $\displaystyle \sin(-\frac{\pi}{4})=-\frac{\sqrt{2}}{2}\ $so
$\displaystyle \arcsin(\sin\frac{5\pi}{4})=\arcsin(-\frac{\sqrt{2}}{2})=-\frac{\pi}{4}$
