# Chapter 1 - Test - Page 46: 22

$\sin\theta = \frac{3}{7}$ (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 $\sqrt 10$ $\sec\theta =\frac{7}{-2\sqrt 10} \times \frac{\sqrt 10}{\sqrt 10} = \frac{-7\sqrt 10}{20}$ $\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{-2\sqrt 10}{7} \times \frac{7}{3} = \frac{-2\sqrt 10}{3}$

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