Question

Find the exact value of each expression:

a) tan^(-1) (1/sqrt3)

b) sec^(-1) (2)

Answer 1

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

Answer 2

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

Answer 3

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 \[tan(\frac{\pi}{6})=\frac{1}{\sqrt{3}}\]

Answer 4

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

Answer 5

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

Answer 6

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