Question 1169243
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Solve the quadratic in *[tex \Large \sec(x)].  Since the angle is obtuse (i.e. the terminal ray is in QII), the cosine and its reciprocal are negative.  Discard the positive root and take the reciprocal of the negative root giving you *[tex \Large \cos(x)].  Then use the Pythagorean identity, *[tex \Large \sin^2(x)\ =\ 1\ -\ \cos^2(x)] to find *[tex \Large \sin(x)] (remember, sine is positive in QII).  Then calculate *[tex \Large \tan(x)\ =\ \frac{\sin(x)}{\cos(x)}]


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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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