How do you determine $tan θ$ $tantheta$ given $cot θ = - \frac{\sqrt{5}}{2}, \frac{π}{2} < θ < π$ $cottheta=-sqrt5/2,pi/2<theta<pi$ ?

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How do you determine $tan θ$ $tantheta$ given $cot θ = - \frac{\sqrt{5}}{2}, \frac{π}{2} < θ < π$ $cottheta=-sqrt5/2,pi/2<theta<pi$ ?

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Answer 1 · 2017-01-24T05:45:43

$tan θ = - \frac{2}{\sqrt{5}}$ $tantheta=-2/sqrt5$ ; $sin θ = - \frac{2}{3}$ $sintheta=-2/3$ ; $csc θ = - \frac{3}{2}$ $csctheta=-3/2$ ; $cos θ = - \frac{\sqrt{5}}{3}$ $costheta=-sqrt5/3$ ; $sec θ = - \frac{3}{\sqrt{5}}$ $sectheta=-3/sqrt5$

Explanation:

Since $tan θ = \frac{1}{cot θ}$ $tantheta=1/cottheta$ , you get:

$tan θ = - \frac{2}{\sqrt{5}}$ $tantheta=-2/sqrt5$

Since $sin θ = \pm \frac{tan θ}{\sqrt{1 + {tan}^{2} θ}}$ $sintheta=+-tantheta/sqrt(1+tan^2theta)$ and sine is positive in the second quadrant:

$sin θ = + \frac{- \frac{2}{\sqrt{5}}}{\sqrt{1 + {(- \frac{2}{\sqrt{5}})}^{2}}}$ $sintheta=+(-2/sqrt5)/sqrt(1+(-2/sqrt5)^2$

$=(-2/sqrt5)/sqrt(1+4/5)=(-2/sqrt5)/sqrt(9/5)=-2/cancelsqrt5*cancelsqrt5/3=-2/3$

Then $csctheta=1/sintheta=-3/2$

$costheta=+-sqrt(1-sin^2theta)$

In the second quadrant cosine is negative, then

$costheta=-sqrt(1-(-2/3)^2)=-sqrt(1-4/9)=-sqrt(5/9)=-sqrt5/3$

and $sectheta=1/costheta=-3/sqrt5$

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How do you determine $tan θ$ $tantheta$ given $cot θ = - \frac{\sqrt{5}}{2}, \frac{π}{2} < θ < π$ $cottheta=-sqrt5/2,pi/2<theta<pi$ ?

1 Answer

Explanation:

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How do you determine $tan θ$ $tantheta$ given $cot θ = - \frac{\sqrt{5}}{2}, \frac{π}{2} < θ < π$ $cottheta=-sqrt5/2,pi/2<theta<pi$ ?