Answer
$\displaystyle \tan t=-\frac{\sqrt{1-\cos^{2}t}}{\cos t}$
Work Step by Step
see: Signs of the Trigonometric Functions (p.412)
$\left[\begin{array}{lll}
Quadrant & Positive & Negative\\
I & all & none\\
II & sin, csc & cos, sec, tan, cot\\
III & tan, cot & sin, csc, cos, sec\\
IV & cos, sec & sin, csc, tan, cot
\end{array}\right]$
also, on p.415, Reciprocal Identities:
$\displaystyle \csc t=\frac{1}{\sin t} \qquad \displaystyle \sec t=\frac{1}{\cos t}\qquad \displaystyle \cot t=\frac{1}{\tan t}$
and: Pythagorean Identities:
$\sin^{2}t+\cos^{2}t=1\qquad\tan^{2}t+1=\sec^{2}t\qquad 1 +\cot^{2}t=\csc^{2}t$
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$\displaystyle \tan t=\frac{\sin t}{\cos t}$, and from
$\sin^{2}t+\cos^{2}t=1\qquad\Rightarrow\sin t=\pm\sqrt{1-\cos^{2}t}$
Determine the sign:
in quadrant III, sine is negative
$\sin t=-\sqrt{1-\cos^{2}t}$
So
$\displaystyle \tan t=\frac{\sin t}{\cos t}$
$\displaystyle \tan t=-\frac{\sqrt{1-\cos^{2}t}}{\cos t}$
(since the denominator, $\cos t$ is negative in q.III,
it will work out for tan to be positive in quadrant III, as the table states)
