If $sin \theta = \frac{1}{9}$, then $tan \theta$ is equal to

 If $sin \theta = \frac{1}{9}$, then $tan \theta$ is equal to

We are given:

$\sin \theta = \frac{1}{9}$

Using the identity:

$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$

Let the opposite side be $1$ and the hypotenuse be $9$.

Using the Pythagorean theorem:

$\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2$

$9^2 = 1^2 + \text{adjacent}^2$

$81 = 1 + \text{adjacent}^2$

$\text{adjacent}^2 = 80$

$\text{adjacent} = \sqrt{80} = 4\sqrt{5}$

Now,

$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{4\sqrt{5}}$

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