Question

What is `cot^2theta` in terms of non-exponential trigonometric functions?

Answer 1

`(1 + cos 2t)/(1 - cos 2t)`

Explanation:

`cot^2 t = (cos^2 t)/(sin^2 t)`
`cos^2 t = (1/2)(1 + cos 2t)`
`sin^2 t = (1/2)(1 - cos 2t)`
Therefor,
`cot^2 t = (1 + cos 2t)/(1 - cos 2t)`