## anonymous 5 years ago Find the exact value of each expression: a) tan^(-1) (1/sqrt3) b) sec^(-1) (2)

1. anonymous

best cheat sheet is here: http://tutorial.math.lamar.edu/cheat_table.aspx

2. anonymous

first one you are looking for a number (angle) between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$whose tangent is $\frac{1}{\sqrt{3}}$

3. anonymous

from the cheat sheet we see that at $\frac{\pi}{6}$ we get $sin(\frac{\pi}{6})=\frac{1}{2}$ and $cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$ and therefore $tan(\frac{\pi}{6})=\frac{1}{\sqrt{3}}$

4. anonymous

so $tan^{-1}(\frac{1}{\sqrt{3}})=\frac{\pi}{6}$

5. anonymous

of course you could just use a calculator. are you working in degrees or radian?

6. anonymous

anyway second one is easier. $sec^{-1}(2)$ is the number whose cosine is $\frac{1}{2}$ and that is $\frac{\pi}{3}$