# How do you evaluate sin(tan^-1(sqrt3))?

$\setminus \frac{\setminus \sqrt{3}}{2}$ , $\setminus \frac{- \setminus \sqrt{3}}{2}$
As you can see from the diagram, $\setminus \theta = \setminus \frac{\setminus \pi}{3} \mathmr{and} \setminus \frac{4 \setminus \pi}{3}$ radians or 60° and 240°
Using the tangent law and sine law of right triangles, you will find that $\setminus \sin \left(\setminus \frac{\setminus \pi}{3}\right) = \setminus \frac{\setminus \sqrt{3}}{2} \mathmr{and} \setminus \sin \left(\setminus \frac{4 \setminus \pi}{3}\right) = - \setminus \frac{\setminus \sqrt{3}}{2}$