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If tan(θ)=-15/8 and sin(θ) is positive find other 5trig ratio and determine which quodrants θ in

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\tan (\theta)=\frac{opposite}{_{\text{ }}adjacent}

Since the sine is positive, we can conclude:

Opposite = 15

Adjacent = -8

Therefore, the angle θ is in the II quadrant, and the other 5 trigonometric functions are given by:


\begin{gathered} \sin (\theta)=\frac{opposite}{_{\text{ }}hypotenuse} \\ _{\text{ }}where\colon \\ _{\text{ }}hypotenuse=\sqrt[]{15^2+(-8)^2} \\ _{\text{ }}hypotenuse=17 \\ so\colon \\ \sin (\theta)=(15)/(17) \end{gathered}
\cos (\theta)=\frac{adjacent}{_{\text{ }}hypotenuse}=-(8)/(17)
\csc (\theta)=\frac{_{\text{ }}hypotenuse}{_{\text{ }}opposite}=(17)/(15)
\sec (\theta)=\frac{_{\text{ }}hypotenuse}{_{\text{ }}adjacent}=-(17)/(8)
\cot (\theta)=\frac{adjacent}{_{\text{ }}opposite}=-(8)/(15)_{}

User AndrewCox
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