Question

How do you evaluate `Tan[arccos ( - sqrt ( 3 ) / 2 ) ]`?

Answer 1

`tan(arccos(-sqrt(3)/2)) = -(sqrt(3))/3`

Explanation:

`tan(theta) = sin(theta)/cos(theta)`
`sin(theta) = +-sqrt(1 - cos^2(theta))`

During the arccosine range, the sine is always positive so

`tan(arccos(-sqrt(3)/2)) = sqrt(1 - (-sqrt(3)/2)^2)/(-sqrt(3)/2)`
`tan(arccos(-sqrt(3)/2)) = -(2*sqrt(1 - 3/4))/sqrt(3)`
`tan(arccos(-sqrt(3)/2)) = -(2*sqrt(3)*sqrt(1 - 3/4))/3`
`tan(arccos(-sqrt(3)/2)) = -(2*sqrt(3)*sqrt((4 - 3)/4))/3`
`tan(arccos(-sqrt(3)/2)) = -(2*sqrt(3)*sqrt(1/4))/3`
`tan(arccos(-sqrt(3)/2)) = -(2*sqrt(3)*(1/2))/3`
`tan(arccos(-sqrt(3)/2)) = -(sqrt(3))/3`