Question 981337
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cot\varphi\ =\ \frac{1}{\tan\varphi}]


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


So, if *[tex \LARGE \tan\theta\ =\ \frac{a}{b}], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\sin\theta\ =\ a\cos\theta]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^2\sin^2\theta\ =\ a^2\ -\ a^2\sin^2\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ \pm\sqrt{\frac{a^2}{a^2\ +\ b^2}}]


Choose the sign based on the given sign relationship.


Similarly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\theta\ =\ \pm\sqrt{\frac{b^2}{a^2\ +\ b^2}}]


Finally,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec\varphi\ =\ \frac{1}{\cos\varphi}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc\varphi\ =\ \frac{1}{\sin\varphi}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \