Question 797740
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\theta\ =\ \frac{1}{3}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pi\ <\ \theta\ <\ \frac{3\pi}{2}\ \Rightarrow\ \theta\ \in \text{Q III}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\varphi\ =\ \frac{\sin\varphi}{\cos\varphi}]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin\theta}{\cos\theta}\ =\ \frac{1}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\sin\theta\ =\ \cos\theta]  


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9\sin^2\theta\ =\ \cos^2\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9\sin^2\theta\ =\ 1\ -\ \sin^2\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\ =\ \frac{1}{10}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ \pm\frac{1}{\sqrt{10}}\ =\ \pm\frac{\sqrt{10}}{10}]


But


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta\ \in \text{Q III}\ \Rightarrow\ \sin\theta\ <\ 0]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ -\frac{\sqrt{10}}{10}]


Since we know


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3\sin\theta\ =\ \cos\theta]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta\ \in \text{Q III}\ \Rightarrow\ \cos\theta\ <\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\theta\ =\ -\frac{3\sqrt{10}}{10}]


And you can finish up for yourself because cotangent is the reciprocal of tangent, secant is the reciprocal of cosine, and cosecant is the reciprocal of sine. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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