Question

How do you find the value of `tan theta` given `sec theta=-9/4` and in quadrant III?

Answer 1

`sqrt65/4`

Explanation:

`cos t = 1/(sec t )= - 4/9`
`sin^2 t = 1 - cos^2 t = 1 - (4/9)^2 = (81 - 16)/81 = 65/81` -->
`sin t = +- sqrt65/9`
Since t is in Quadrant III, then sin t is negative.
`tan t = sint/(cos t) = (- sqrt65/9)(-9/4) = sqrt65/4`