Answer
(a) $30^\circ$
(b) $-45^\circ$
(c) $135^\circ$
(d) $-60^\circ$
Work Step by Step
(a) Since $cos(30^\circ)=\frac{\sqrt 3}{2}$, we have $\theta=arccos(\frac{\sqrt 3}{2})=30^\circ$
(b) Since $tan(-45^\circ)=-1$, we have $\theta=tan^{-1}(-1)=-45^\circ$
(c) Since $cot(135^\circ)=-1$, we have $\theta=cot^{-1}(-1)=135^\circ$
(d) Since $csc(-60^\circ)=-\frac{2\sqrt 3}{3}$, we have $\theta=csc^{-1}(-\frac{2\sqrt 3}{3})=-60^\circ$
