Answer
a) $\csc^{-1}\sqrt2=\frac{\pi}{4}$
b) $\arcsin1=\frac{\pi}{2}$
Work Step by Step
a) let $\csc^{-1}\sqrt2=\theta, -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}, \theta\neq0$
$\csc\theta=\sqrt2$
$\sin\theta=\frac{1}{\sqrt2}$
$\theta=\frac{\pi}{4}+2k\pi, \frac{3\pi}{4}+2k\pi$ where k is an integer
$\because -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}, \theta\neq0$, let $k=0, \theta=\frac{\pi}{4}$
$\therefore \csc^{-1}\sqrt2=\frac{\pi}{4}$
b) let $\arcsin1=\theta, -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$
$\sin\theta=1$
$\theta=\frac{\pi}{2}+2k\pi$ where k is an integer
$\because -\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$, let $k=0, \theta=\frac{\pi}{2}$
$\therefore \arcsin1=\frac{\pi}{2}$
