Question 616932
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2.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(A)\ -\ \sec(A)\ =\ \frac{\sin(A)}{\cos(A)}\ -\ \frac{1}{\cos(A)}\ =\ \frac{sin(A)\ -\ 1}{\cos(A)}]


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