Question

How do you calculate `tan(arccos(-9/10))`?

Answer 1

In this way.

We have to calculate `tanalpha`, where alpha is:

`alpha=arccos(-9/10)`.

Since the range of the function `y=arccosx` is `[0,pi]` and the value negative `-9/10`, the angle is in the second quadrant, in which the sinus is positive. So:

`cosalpha=-9/10`,

`sinalpha=+sqrt(1-cos^2alpha)=+sqrt(1-81/100)=sqrt19/10`.

Than:

`tanalpha=sinalpha/cosalpha=(sqrt19/10)/(-9/10)=-sqrt19/10*10/9=`

`=-sqrt19/9`.