Answer
$\sin\theta=\dfrac{12}{13}$
$\cos\theta=-\dfrac{5}{13}$
$\cot\theta=-\dfrac{5}{12}$
$\sec\theta=-\dfrac{13}{5}$
$\csc\theta=\dfrac{13}{12}$
Work Step by Step
Determine $\sec\theta$:
$\tan^2\theta+1=\sec^2\theta$
$\left(-\dfrac{12}{5}\right)^2+1=\sec^2\theta$
$\sec\theta=\pm\sqrt{\dfrac{169}{25}}=\pm\dfrac{13}{5}$
Because the angle $\theta$ is in Quadrant II, we have:
$\sec\theta=-\dfrac{13}{5}$
Determine $\cos\theta,\sin\theta,\cot\theta,\csc\theta$:
$\cos\theta=\dfrac{1}{\sec\theta}=\dfrac{1}{-\frac{13}{5}}=-\dfrac{5}{13}$
$\sin\theta=\cos\theta\cdot\tan\theta=-\dfrac{5}{13}\cdot \left(-\dfrac{12}{5}\right)=\dfrac{12}{13}$
$\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{1}{-\frac{12}{5}}=-\dfrac{5}{12}$
$\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{1}{\frac{12}{13}}=\dfrac{13}{12}$
