To find the tangent of an angle θ in Quadrant II, where \(\sin(\theta) = \frac{1}{4}\), we can use the Pythagorean identity, which states:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
First, we calculate \(\sin^2(\theta)\):
\[ \sin^2(\theta) = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]
Now, we can plug this value into the Pythagorean identity:
\[ \frac{1}{16} + \cos^2(\theta) = 1 \]
To isolate \(\cos^2(\theta)\), we subtract \(\frac{1}{16}\) from both sides:
\[ \cos^2(\theta) = 1 - \frac{1}{16} \]
To perform the subtraction, we convert \(1\) to a fraction with a denominator of \(16\):
\[ 1 = \frac{16}{16} \]
Therefore,
\[ \cos^2(\theta) = \frac{16}{16} - \frac{1}{16} = \frac{15}{16} \]
Now, we take the square root to find \(\cos(\theta)\):
\[ \cos(\theta) = \pm \sqrt{\frac{15}{16}} = \pm \frac{\sqrt{15}}{4} \]
Since θ is in Quadrant II, where the cosine function is negative, we have:
\[ \cos(\theta) = -\frac{\sqrt{15}}{4} \]
Next, we can find \(\tan(\theta)\) using the identity:
\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \]
Substituting the known values:
\[ \tan(\theta) = \frac{\frac{1}{4}}{-\frac{\sqrt{15}}{4}} = \frac{1}{4} \cdot -\frac{4}{\sqrt{15}} = -\frac{1}{\sqrt{15}} \]
Now to simplify it further, we can rationalize the denominator:
\[ \tan(\theta) = -\frac{1}{\sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}} = -\frac{\sqrt{15}}{15} \]
Thus, the tangent of the angle θ is:
\[ \tan(\theta) = -\frac{\sqrt{15}}{15} \]
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