Question 1026524
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan\varphi\ =\ \frac{\sin\varphi}{\cos\varphi}]


so


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


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\varphi\ +\ \sin^2\varphi\ =\ 1]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\varphi\ -\ 1\ =\ -\cos^2\varphi]


That should give you enough information to solve the first one.


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

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

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