Question

If `cos theta = -8/9`, and `theta` is in Quadrant III, how do you find `tanthetacottheta+csctheta`?

Answer 1

See below. Answer: `(17-9sqrt(17))/17`

Explanation:

Original question: Given `costheta=-8/9`, `theta` in Quadrant III, find `tanthetacottheta+csctheta`

First, we notice that the question is really only asking us to find `sintheta` since `tanthetacottheta+csctheta=(sintheta/costheta)(costheta/sintheta)+1/sintheta=1+1/sintheta`, making this problem a lot less complicated than it seems.

We also know that since `theta` is in Quadrant III, both `costheta` and `sintheta` are negative.

Consider Pythagorean's identity, `sin^2theta+cos^2theta=1`.

Since we know that `cos(theta)=-8/9`, we can plug in this value into Pythagorean's identity to find `sin(theta)`:
`sin^2theta+(-8/9)^2=1`
`sin^2theta+64/81=1`
`sin^2theta=17/81`
`sintheta=+-sqrt(17/81)=+-sqrt(17)/9`
However, since we know that `sintheta` is in Quadrant III and must be negative, then `sintheta=-sqrt(17)/9`

Finally, since we are solving for `1+csctheta` or `1+1/sintheta`, our final answer is `1+1/(-sqrt(17)/9)=1-9/sqrt(17)=(17-9sqrt(17))/17`.

Answer: `(17-9sqrt(17))/17`