$Find the angle where the tangent is equal to frac{1}{sqrt{3}} on the unit circle.$

To find the angle where the tangent is equal to \( \frac{1}{\sqrt{3}} \) on the unit circle, we need to find the angles θ that satisfy this condition.

From trigonometric identities, we know that:

$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$

Given:

$\tan(\theta) = \frac{1}{\sqrt{3}}$

We recognize that:

$\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$

Since the tangent function has a period of \( \pi \), the general solution for θ is:

$\theta = \frac{\pi}{6} + k\pi\ (k \in \mathbb{Z})$

We need to determine where ( an( heta) = frac{1}{sqrt{3}} ) on the unit circle.

We know:

$ an( heta) = frac{sin( heta)}{cos( heta)}$

Given:

$ an( heta) = frac{1}{sqrt{3}}$

We identify that this is true for:

$ heta = frac{pi}{6}$

And since the tangent function repeats every π, the general solution is:

$ heta = frac{pi}{6} + npi (n in mathbb{Z})$

For ( an( heta) = frac{1}{sqrt{3}} ),

we find:

$ heta = frac{pi}{6}$

And the general solution is:

$ heta = frac{pi}{6} + kpi (k in mathbb{Z})$

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