Question

How do you evaluate `tan(sec^-1 4)`?

Answer 1

`tan(sec^-1 4)=sqrt15`

Explanation:

`tan(sec^-1 4)`

The restriction for the range of `sec^-1x` is `[0,pi],y!=pi/2` and since the argument is positive it means that our triangle is in quadrant I with the hypotenuse of 4 and adjacent of 1 and therefore using pythagorean theorem we have the opposite is `sqrt15`.

Therefore, the ratio for tangent from our triangle is
`tan theta=o/a=sqrt15/1=sqrt15`