Question

Given that `csctheta=-8/5` where `pi<theta<(3pi)/2`, how do you determine the exact value of `tantheta`?

Answer 1

`tan theta = 5/sqrt39`

Explanation:

`csc theta = -8/5 :. sin theta = -5/8 ; pi < theta <(3pi)/2, theta` is in third quadrant. Let `P =`perpendicular of rt triangle , `H=`hypotenuse of a rt. triangle and `B=` base of rt.triangle.

`Sin theta = P/H :. P = -5 ; H=8 :. B= sqrt(H^2-P^2)= sqrt(64-25)=+-sqrt39 = -sqrt39` In `3` rd quadrant both `P and B` are negative.

`:. tan theta = P/B = (-5)/-sqrt39 = 5/sqrt39` [Ans]