Answer
a) $\frac{\pi}{4}$
b) $\frac{5\pi}{6}$
c) $0$
d) $-1/2$
Work Step by Step
a) First, we look at the interval of the inverse of tangent which is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. Then, we must find what angle will give a slope of $1$ in that interval. The answer is $\frac{\pi}{4}$
b) First, we look at the interval of the inverse of cosine which is $0\lt x\lt \pi$. Then, we must find what angle will give a value of $-\sqrt3/2$ in that interval. The answer is $\frac{5\pi}{6}$
c) First, we look at the interval of the inverse of tangent which is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. Since the tangents cancel out, we must find what angle is symmetrical to $3\pi$ in that interval and that will be $0$
d) First, we'll calculate tan$^-1(-\sqrt3)$ whose interval is $-\frac{\pi}{2}\lt x\lt\frac{\pi}{2}$. Then, we must find what angle will give a slope of $-\sqrt3$ in that interval. That will be $-\frac{\pi}{3}$.
Now, we can calculate cos($-\frac{\pi}{3})$ which equals -1/2
