Question 997984
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan2x\ =^?\ \frac{-2\tan x}{\sec^2x\ -\ 2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =^?\ \frac{\frac{-2\sin x}{\cos x}}{\frac{1}{\cos^2x}\ -\ \frac{2\cos^2x}{\cos^2x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =^?\ \left(\frac{-2\sin x}{\cos x}\right)\left(\frac{\cos^2x}{1\ -\ 2\cos^2x}\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =^?\ \frac{-2\sin x\cos x}{1\ -\ 2\cos^2x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =^?\ \frac{\sin2x}{\cos2x}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv\ \tan2x]



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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \