Question

How do you evaluate `cot[arccos (-4/5)]`?

Answer 1

`cot[arccos(-4/5)]=-4/3`

Explanation:

`cot[arccos(-4/5)]`-->restriction for `arccos x` is `[0,pi]`.

Since the argument is negative that means our triangle is in quadrant two with the adjacent side -4 and hypotenuse 5. Hence this is a 3-4-5 triangle with the opposite side being 3.

Therefore `cot theta` = adjacent/opposite = `a/o = -4/3`