Question 984348
<font face="Times New Roman" size="+2">


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2\varphi\ =\ \frac{a^2}{b^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \cos^2\varphi\ =\ \frac{a^2}{b^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\varphi\ =\ \frac{b^2\ -\ a^2}{b^2}]


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


Choose the sign based on the quadrant given in the original problem.


Then


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


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec\varphi\ =\ \frac{1}{\cos\varphi}]


While you are at it, you might make note of the facts that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cot\varphi\ =\ \frac{1}{\tan\varphi}\ =\ \frac{\cos\varphi}{\sin\varphi}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \csc\varphi\ =\ \frac{1}{\sin\varphi}] 


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

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