Answer
$\sin\theta =-\displaystyle \frac{12}{13}$
$\cos\theta =\displaystyle \frac{5}{13}$
$\tan\theta =-\displaystyle \frac{12}{5}$
$\cot\theta =-\displaystyle \frac{5}{12}$
$\sec\theta =\displaystyle \frac{13}{5}$
$\csc\theta =-\displaystyle \frac{13}{12}$
Work Step by Step
For any real number $s$ represented by a directed arc on the unit circle,
$\sin s=y\quad \cos s=x \quad \displaystyle \tan s=\frac{y}{x} (x\neq 0)$
$\displaystyle \csc s=\frac{1}{y} (y\neq 0)\quad \displaystyle \sec s=\frac{1}{x} (x\neq 0) \displaystyle \quad\cot s=\frac{x}{y} (y\neq 0)$.
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$\displaystyle \sin\theta=y=-\frac{12}{13}$
$\displaystyle \cos\theta=x=\frac{5}{13}$
$\displaystyle \tan\theta=\frac{y}{x}=\frac{-\frac{12}{13}}{\frac{5}{13}}=-\frac{12}{13}(\frac{13}{5})=-\frac{12}{5}$
$\displaystyle \cot\theta=\frac{x}{y}=\frac{\frac{5}{13}}{-\frac{12}{13}}=\frac{5}{13}(-\frac{13}{12})=-\frac{5}{12}$
$\displaystyle \sec\theta=\frac{1}{x}=\frac{1}{\frac{5}{13}}=\frac{13}{5}$
$\displaystyle \csc\theta=\frac{1}{y}=\frac{1}{-\frac{12}{13}}=-\frac{13}{12}$