To find the tangent of the angle given that \(\sin(\theta) = \frac{\sqrt{5}}{5}\) and that \(\theta\) is in quadrant II, we can use the Pythagorean identity and the properties of sine and cosine in that quadrant.
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Use the Pythagorean identity: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \] Plug in the value of sine: \[ \left(\frac{\sqrt{5}}{5}\right)^2 + \cos^2(\theta) = 1 \] This simplifies to: \[ \frac{5}{25} + \cos^2(\theta) = 1 \] \[ \frac{1}{5} + \cos^2(\theta) = 1 \] Now, isolate \(\cos^2(\theta)\): \[ \cos^2(\theta) = 1 - \frac{1}{5} = \frac{4}{5} \]
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Find \(\cos(\theta)\): Since \(\theta\) is in quadrant II, cosine is negative: \[ \cos(\theta) = -\sqrt{\frac{4}{5}} = -\frac{2}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \]
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Calculate \(\tan(\theta)\) using the tangent identity: \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \] Substitute the values we have: \[ \tan(\theta) = \frac{\frac{\sqrt{5}}{5}}{-\frac{2\sqrt{5}}{5}} = \frac{\sqrt{5}}{5} \cdot \frac{-5}{2\sqrt{5}} \] Simplifying this gives the following: \[ = \frac{-1}{2} \]
Thus, the tangent of the angle \(\theta\) is: \[ \boxed{-\frac{1}{2}} \]
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