Question 497100
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If *[tex \Large cos\theta\ =\ \frac{-12}{13}] then *[tex \Large cos^2\theta\ =\ \frac{144}{169].


Since we know *[tex \Large cos^2\varphi\ +\ sin^2\varphi\ =\ 1], we can determine that *[tex \Large sin^2\theta\ =\ 1\ -\ \frac{144}{169}\ =\ \frac{25}{169}]


and therefore *[tex \Large \sin\theta\ =\ \frac{\pm{5}}{13}], taking the positive side because sin is positive in QII, we have *[tex \Large \sin\theta\ =\ \frac{5}{13}]


From there, the rest are easy:  *[tex \Large \tan\varphi\ =\ \frac{\sin\varphi}{\cos\varphi}], *[tex \Large \cot\varphi\ =\ \frac{1}{tan\varphi}\ =\ \frac{\cos\varphi}{\sin\varphi}], *[tex \Large \sec\varphi\ =\ \frac{1}{\cos\varphi}], and *[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
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