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

Jun 7, 2017

$\tan \theta = \frac{5}{\sqrt{39}}$
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 $P \mathmr{and} B$ are negative.
$\therefore \tan \theta = \frac{P}{B} = \frac{- 5}{-} \sqrt{39} = \frac{5}{\sqrt{39}}$ [Ans]