Question 778054
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Pythagorean Identity: *[tex \LARGE \sin^2\theta\ +\ \cos^2\theta\ =\ 1]


Divide both sides by *[tex \LARGE \cos^2\theta]


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


A little algebra music:


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


Since *[tex \LARGE \tan\varphi\ =\ \frac{\sin\varphi}{\cos\varphi}] and *[tex \LARGE \sec\varphi\ =\ \frac{1}{\cos\varphi}], just make the substitutions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan^2\theta\ +\ 1\ =\ \sec^2\theta]


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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