Answer
$\sin\theta = \frac{3}{7}$ (Given)
$\cos\theta = \frac{-2\sqrt 10}{7}$
$\tan\theta = \frac{-3\sqrt 10}{20}$
$\csc\theta = \frac{7}{3}$
$\sec\theta = \frac{-7\sqrt 10}{20}$
$\cot\theta = \frac{-2\sqrt 10}{3}$
Work Step by Step
Given that $\sin\theta = \frac{3}{7}$ and $\theta$ is in quadrant II
Using the Identity
$\sin^{2}\theta + \cos^{2}\theta = 1$
$(\frac{3}{7})^{2} + \cos^{2}\theta = 1$
$(\frac{9}{49})+ \cos^{2}\theta = 1$
$ \cos^{2}\theta = 1 - (\frac{9}{49})$
$ \cos^{2}\theta = \frac{49-9}{49}$
$ \cos^{2}\theta = \frac{40}{49}$
$ \cos\theta = +\frac{\sqrt 40}{7}$ or $ \cos\theta = -\frac{\sqrt 40}{7}$
In quadrant II, $\cos\theta$ is negative. So,
$ \cos\theta = -\frac{\sqrt 40}{7} = -\frac{\sqrt (4\times 10)}{7} = \frac{-2\sqrt 10}{7}$
$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{3}{7}}{\frac{-2\sqrt 10}{7}} = \frac{3}{7} \times \frac{7}{-2\sqrt 10} = \frac{-3}{2\sqrt 10}$
To rationalize the denominator, multiply and divide by $\sqrt 10$
$\tan\theta = \frac{-3}{2\sqrt 10} \times \frac{\sqrt 10}{\sqrt 10} = \frac{-3\sqrt 10}{20}$
$\csc\theta = \frac{1}{\sin\theta} = \frac{1}{\frac{3}{7}} = \frac{7}{3}$
$\sec\theta = \frac{1}{\cos\theta} = \frac{1}{\frac{-2\sqrt 10}{7}} = \frac{7}{-2\sqrt 10}$
To rationalize the denominator, multiply and divide by $\sqrt 10$
$\sec\theta =\frac{7}{-2\sqrt 10} \times \frac{\sqrt 10}{\sqrt 10} = \frac{-7\sqrt 10}{20}$
$\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{-2\sqrt 10}{7} \times \frac{7}{3} = \frac{-2\sqrt 10}{3}$
