Question 982115
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan A\ =\ \frac{\sin A}{\cos A}]


by definition of the tangent function.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan A\ =\ \frac{a}{b}]


is given.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{a}{b}\ =\ \frac{\sin A}{\cos A}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\cos A\ =\ b\sin A]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\cos^2A\ =\ b^2\sin^2A]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\cos^2A\ =\ b^2\left(1\ -\ \cos^2A\right)]


by the Pythagorean Identity


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2\cos^2A\ +\ b^2\cos^2A\ =\ b^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(a^2\ +\ b^2\right)\cos^2\ =\ b^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2A\ =\ \frac{b^2}{a^2\ +\ b^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos A\ =\ \pm\sqrt{\frac{b^2}{a^2\ +\ b^2}}]


Since *[tex \Large 0\ <\ A\ \frac{\pi}{2}], *[tex \Large \cos A\ >\ 0]


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos A\ =\ \sqrt{\frac{b^2}{a^2\ +\ b^2}}]


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

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