Question 879366
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos\theta\ =\ \frac{12}{13}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\theta\ =\ \frac{144}{169}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \cos^2\theta\ =\ \frac{25}{169}]


But


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


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\theta\ =\ \frac{25}{169}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ \pm\frac{5}{13}]


Then, since *[tex \Large \tan\varphi\ =\ \frac{\sin\varphi}{\cos\varphi}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\theta\ =\ \frac{\sin\theta}{\cos\theta}\ =\ \frac{\pm\frac{5}{13}}{\frac{12}{13}}\ =\ \pm\frac{5}{12}]


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