Question 465237
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Given:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ =\ \frac{3}{4}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec(x)\ <\ 0]


But:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sec(x)\ <\ 0\ \Leftrightarrow\ \frac{1}{\sec(x)}\ <\ 0\ \Leftrightarrow\ \cos(x)\ <\ 0]


Since:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin(x)\ =\ \frac{3}{4}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(x)\ =\ \frac{9}{16}]


And since:


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


We can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2(x)\ =\ \frac{7}{16}]


Which, coupled with the fact that *[tex \Large \cos(x)\ <\ 0], leads us to deduce that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(x)\ =\ -\frac{\sqrt{7}}{4}]


Then, since we know that *[tex \Large \tan(\varphi)\ =\ \frac{\sin(\varphi)}{\cos(\varphi)}], we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(x)\ =\ \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}}\ =\ \frac{3\sqrt{7}}{7}]


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