Question 992389
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Since secant is negative (given) and we know that cosine is the reciprocal of secant, then cosine must be negative as well.  You have the correct magnitude for cosine, but your sign is incorrect.  When you solved for the value of cosine, you had to take the square root of cosine squared.  When you take the square root of both sides of an equation you have to consider both possible signs for the result.  That is why the problem specified a negative secant -- so that you would know which sign to attach to cosine at the point where you took the square root.


Since sine is positive and cosine is negative, we know from the identity *[tex \Large \tan\varphi\ =\ \frac{\sin\varphi}{\cos\varphi}] that tangent must be negative, and therefore cotangent, the reciprocal of tangent must be negative as well.


Your value for cotangent is incorrect.  Sine over cosine is tangent, so *[tex \Large \tan(t)\ =\ -\frac{1}{2\sqrt{6}}], and therefore *[tex \Large \cot(t)\ =\ -2\sqrt{6}]


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

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