Answer: Choice B
[tex]-\frac{\sqrt{91}}{3}[/tex]
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Explanation:
We're given [tex]\cos(\theta) = -\frac{3}{10}[/tex]
Use that value of cosine to find sine.
[tex]\sin^2(\theta) + \cos^2(\theta) = 1\\\\\sin^2(\theta) = 1-\cos^2(\theta)\\\\\sin(\theta) = \sqrt{1-\cos^2(\theta)} \ \text{ ... sine is positive in Q2}\\\\\sin(\theta) = \sqrt{1-\left(-\frac{3}{10}\right)^2}\\\\\sin(\theta) = \sqrt{1-\frac{9}{100}}\\\\\sin(\theta) = \sqrt{\frac{100}{100}-\frac{9}{100}}\\\\\sin(\theta) = \sqrt{\frac{100-9}{100}}\\\\\sin(\theta) = \sqrt{\frac{91}{100}}\\\\\sin(\theta) = \frac{\sqrt{91}}{\sqrt{100}}\\\\\sin(\theta) = \frac{\sqrt{91}}{10}\\\\[/tex]
Now divide the sine and cosine values to find tangent.
[tex]\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\\\\\tan(\theta) = \sin(\theta) \div \cos(\theta)\\\\\tan(\theta) = \frac{\sqrt{91}}{10} \div -\frac{3}{10}\\\\\tan(\theta) = \frac{\sqrt{91}}{10} \times -\frac{10}{3}\\\\\tan(\theta) = -\frac{\sqrt{91}}{3}\\\\[/tex]
Tangent is always negative in Q2.
