If sin theta is equal to 2/3, theta not in quadrant 1, find tan theta?

Here,

So, we have

Now , using trig.identity , we get

Hence,

Given that $\sin(\theta) = \frac{2}{3}$ and $\theta$ is not in quadrant 1, we can use the relationship between sine and cosine in different quadrants to determine the sign of $\cos(\theta)$. Since $\sin(\theta) = \frac{2}{3}$ is positive, $\cos(\theta)$ must also be positive because in quadrants where sine is positive, cosine is also positive.

Using the Pythagorean identity $\sin^2(\theta) + \cos^2(\theta) = 1$, we can solve for $\cos(\theta)$ to find $\cos(\theta) = \sqrt{1 - \sin^2(\theta)}$. Then, we can find $\tan(\theta)$ using the relationship $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.

Substituting the given values, we have $\tan(\theta) = \frac{\frac{2}{3}}{\sqrt{1 - \left(\frac{2}{3}\right)^2}}$. After simplification, we get $\tan(\theta) = \frac{2}{\sqrt{5}}$. Therefore, $\tan(\theta) = \frac{2\sqrt{5}}{5}$ because $\sqrt{5} \times \sqrt{5} = 5$.