Answer
$\cos\theta=-\dfrac{2\sqrt 6}{7}$
$\tan\theta=-\dfrac{5\sqrt 6}{12}$
$\cot\theta=-\dfrac{2\sqrt 6}{5}$
$\sec\theta=-\dfrac{7\sqrt 6}{12}$
$\csc\theta=\dfrac{7}{5}$
Work Step by Step
Determine $\cos\theta$:
$\sin^2\theta+\cos^2\theta=1$
$\cos\theta=\pm\sqrt{1-\sin^2\theta}=\pm\sqrt{1-\left(\dfrac{5}{7}\right)^2}$
$=\pm\sqrt{\dfrac{24}{49}}$
$=\pm\dfrac{2\sqrt 6}{7}$
Because the angle $\theta$ is in Quadrant II, we have:
$\cos\theta=-\dfrac{2\sqrt 6}{7}$
Determine $\tan\theta,\cot\theta,\sec\theta,\csc\theta$:
$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\frac{5}{7}}{-\frac{2\sqrt 6}{7}}=-\dfrac{5}{2\sqrt 6}=-\dfrac{5\sqrt 6}{12}$
$\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{1}{-\frac{5}{2\sqrt 6}}=-\frac{5}{2\sqrt 6}=-\dfrac{2\sqrt 6}{5}$
$\sec\theta=\dfrac{1}{\cos\theta}=\dfrac{1}{-\frac{2\sqrt 6}{7}}=-\dfrac{7}{2\sqrt 6}=-\dfrac{7\sqrt 6}{12}$
$\csc\theta=\dfrac{1}{\sin\theta}=\dfrac{1}{\frac{5}{7}}=\dfrac{7}{5}$
