How do you find the five remaining trigonometric function satisfying $csc θ = - \frac{3}{2}$ $csctheta=-3/2$ , $cos θ < 0$ $costheta<0$ ?

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How do you find the five remaining trigonometric function satisfying $csc θ = - \frac{3}{2}$ $csctheta=-3/2$ , $cos θ < 0$ $costheta<0$ ?

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Answer 1 · 2017-02-01T13:17:41

$csc t = \frac{1}{sin t} = - \frac{3}{2}$ $csc t = 1/(sin t) = - 3/2$ --> $sin t = - \frac{2}{3}$ $sin t = - 2/3$
$sin t = - \frac{2}{3}$ $sin t = - 2/3$
${cos}^{2} t = 1 - {sin}^{2} t = 1 - \frac{4}{9} = \frac{5}{9}$ $cos^2 t = 1 - sin^2 t = 1 - 4/9 = 5/9$ --> $cos t = \pm \frac{\sqrt{5}}{3}$ $cos t = +- sqrt5/3$
$cos t = - \frac{\sqrt{5}}{3}$ $cos t = - sqrt5/3$ (cos t < 0)
$tan t = \frac{sin}{cos} = (- \frac{2}{3}) (- \frac{3}{\sqrt{5}}) = \frac{2}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}$ $tan t = sin/(cos) = (- 2/3)(- 3/sqrt5) = 2/sqrt5 = (2sqrt5)/5$
$cot t = \frac{1}{tan t} = \frac{5}{2 \sqrt{5}}$ $cot t = 1/(tan t) = 5/(2sqrt5)$ .
$sec t = \frac{1}{cos} = - \frac{3}{\sqrt{5}} = - \frac{3 \sqrt{5}}{5}$ $sec t = 1/(cos) = - 3/sqrt5 = - (3sqrt5)/5$ .
$csc t = - \frac{3}{2}$ $csc t = -3/2$

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How do you find the five remaining trigonometric function satisfying $csc θ = - \frac{3}{2}$ $csctheta=-3/2$ , $cos θ < 0$ $costheta<0$ ?

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