To solve for [tex]\(\tan(\theta)\)[/tex] given [tex]\(\sin(\theta) = \frac{\sqrt{37}}{37}\)[/tex] and [tex]\(\theta\)[/tex] is in quadrant II, follow these steps:
1. Identify [tex]\(\sin(\theta)\)[/tex]:
[tex]\[
\sin(\theta) = \frac{\sqrt{37}}{37}
\][/tex]
2. Recognize that [tex]\(\theta\)[/tex] is in quadrant II:
In quadrant II, the sine function is positive, and the cosine function is negative.
3. Use the Pythagorean identity to find [tex]\(\cos(\theta)\)[/tex]:
The Pythagorean identity states:
[tex]\[
\sin^2(\theta) + \cos^2(\theta) = 1
\][/tex]
Substituting [tex]\(\sin(\theta)\)[/tex]:
[tex]\[
\left(\frac{\sqrt{37}}{37}\right)^2 + \cos^2(\theta) = 1
\][/tex]
Simplifying the square:
[tex]\[
\frac{37}{1369} + \cos^2(\theta) = 1
\][/tex]
Solving for [tex]\(\cos^2(\theta)\)[/tex]:
[tex]\[
\cos^2(\theta) = 1 - \frac{37}{1369} = \frac{1369}{1369} - \frac{37}{1369} = \frac{1332}{1369}
\][/tex]
Therefore:
[tex]\[
\cos(\theta) = \pm \sqrt{\frac{1332}{1369}} = \pm \frac{\sqrt{1332}}{37}
\][/tex]
Since [tex]\(\theta\)[/tex] is in quadrant II, [tex]\(\cos(\theta)\)[/tex] is negative:
[tex]\[
\cos(\theta) = -\frac{\sqrt{1332}}{37}
\][/tex]
4. Simplify [tex]\(\sqrt{1332}\)[/tex]:
[tex]\(\sqrt{1332}\)[/tex] can be simplified, but we have calculated it for us and know:
[tex]\[
\cos(\theta) \approx -0.9863939238321437
\][/tex]
5. Calculate [tex]\(\tan(\theta)\)[/tex]:
The tangent function is defined as the ratio of sine to cosine:
[tex]\[
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{\sqrt{37}}{37}}{-\frac{\sqrt{1332}}{37}} = -\frac{\sqrt{37}}{\sqrt{1332}}
\][/tex]
Simplifying this fraction gives:
[tex]\[
\tan(\theta) \approx -0.16666666666666666
\][/tex]
Converting that to a fraction, we get:
[tex]\[
\tan(\theta) = -\frac{1}{6}
\][/tex]
Thus, the correct answer is:
[tex]\[
\boxed{-\frac{1}{6}}
\][/tex]
