If sin(θ)=2/3, and θ is in quadrant II , then find: tan(θ)


To find \(\tan(\theta)\) given that \(\sin(\theta) = \frac{2}{3}\) and \(\theta\) is in quadrant II, follow these steps:

1. Recall the identity for \(\sin(\theta)\):
   \[
   \sin^2(\theta) + \cos^2(\theta) = 1
   \]
   Since \(\sin(\theta) = \frac{2}{3}\), substitute this into the equation:
   \[
   \left(\frac{2}{3}\right)^2 + \cos^2(\theta) = 1
   \]
   \[
   \frac{4}{9} + \cos^2(\theta) = 1
   \]
2. Solve for \(\cos^2(\theta)\):
   \[
   \cos^2(\theta) = 1 - \frac{4}{9} = \frac{9}{9} - \frac{4}{9} = \frac{5}{9}
   \]
3. Take the square root of both sides to find \(\cos(\theta)\):
   \[
   \cos(\theta) = \pm \frac{\sqrt{5}}{3}
   \]
   Since \(\theta\) is in quadrant II, \(\cos(\theta)\) is negative:
   \[
   \cos(\theta) = -\frac{\sqrt{5}}{3}
   \]
4. Now, use the identity for \(\tan(\theta)\):
   \[
   \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
   \]
   Substitute the values of \(\sin(\theta)\) and \(\cos(\theta)\):
   \[
   \tan(\theta) = \frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}} = -\frac{2}{\sqrt{5}}
   \]
   To rationalize the denominator:
   \[
   \tan(\theta) = -\frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5}
   \]

Thus, \(\tan(\theta) = -\frac{2\sqrt{5}}{5}\).

Would you like more details or have any questions?

Here are five related questions:
1. How do you find the values of \(\cos(\theta)\) and \(\tan(\theta)\) in other quadrants?
2. What are the steps to solve trigonometric identities involving \(\sin\), \(\cos\), and \(\tan\)?
3. How do you determine the signs of trigonometric functions in each quadrant?
4. Can you prove that \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\)?
5. How do you simplify trigonometric expressions with irrational denominators?

**Tip:** Always remember the ASTC rule (All Students Take Calculus) to determine the signs of trigonometric functions in different quadrants.

If sin(θ) = 2/3, and θ is in quadrant II, then find: tan(θ)

Learn how to find tan(θ) when sin(θ) = 2/3 and θ is in quadrant II. This step-by-step solution uses the Pythagorean identity and trigonometric function relationships. Suitable for students in Grades 10-12.

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