Question 193369
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I'll do you one better.


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


To begin, we know:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}] and *[tex \LARGE \sec{\theta} = \frac{1}{\cos{\theta}}]


So:


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


But:


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


But:


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


So:


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


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1 + \tan^2{\theta}\equiv\sec^2{\theta}]


for all real values of *[tex \LARGE \theta]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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