Question 454486
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Recall that secant is one over cosine and tangent is sine over cosine, so your problem becomes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\cos\theta}\ -\ \frac{\sin\theta}{\cos\theta}\ =\ 0]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sin\theta}{\cos\theta}\ =\ \frac{1}{\cos\theta}]


Multiply by *[tex \LARGE \cos\theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\theta\ =\ 1]


Check the unit circle, recalling that the sin value is the *[tex \Large y]-coordinate


<img src="http://www.math.ucsd.edu/~jarmel/math4c/Unit_Circle_Angles.png">


to see that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta\ =\ \sin^{-1}(1)\ =\ \frac{\pi}{2}]


But recall that you can go around the circle as many times as you like since you didn't provide a range for the answer and that once around is *[tex \Large 2\pi]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta\ =\ \frac{\pi}{2}\ +\ 2k\pi]


where *[tex \LARGE k\ \in\ \mathbb{Z}]


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