Given that $csc θ = - \frac{8}{5}$ $csctheta=-8/5$ where $π < θ < \frac{3 π}{2}$ $pi<theta<(3pi)/2$ , how do you determine the exact value of $tan θ$ $tantheta$ ?

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Given that $csc θ = - \frac{8}{5}$ $csctheta=-8/5$ where $π < θ < \frac{3 π}{2}$ $pi<theta<(3pi)/2$ , how do you determine the exact value of $tan θ$ $tantheta$ ?

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Answer 1 · 2017-06-07T10:42:23

$tan θ = \frac{5}{\sqrt{39}}$ $tan theta = 5/sqrt39$

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]

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Given that $csc θ = - \frac{8}{5}$ $csctheta=-8/5$ where $π < θ < \frac{3 π}{2}$ $pi<theta<(3pi)/2$ , how do you determine the exact value of $tan θ$ $tantheta$ ?

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Given that $csc θ = - \frac{8}{5}$ $csctheta=-8/5$ where $π < θ < \frac{3 π}{2}$ $pi<theta<(3pi)/2$ , how do you determine the exact value of $tan θ$ $tantheta$ ?