There are 2 ways to do this:
1) Transform tan into terms of sin.
\(tan x = \frac{sin x}{cos x} = \frac{sin x}{\sqrt{1-sin^2 x}}\)
where \(sin x = sin(sin^{-1} (\frac{2}{5})) = \frac{2}{5}\)
Substituting back in gives:
\(tan x = \frac{\frac{2}{5}}{\sqrt{1-(\frac{2}{5})^2}} = \frac{\frac{2}{5}}{\sqrt{\frac{21}{25}}} = \frac{2}{5}*\frac{\sqrt{25}}{\sqrt{21}} = \frac{2}{\sqrt{21}}\)
2) Use a right triangle.
\(\theta = sin^{-1} (\frac{2}{5}) \\ \\ sin \theta = \frac{2}{5}\)
sin = opp/hyp --> opp = 2, hyp = 5
Use Pythagorean theorem to solve for adjacent side.
\(adj = \sqrt{5^2 - 2^2} = \sqrt{21}\)
tan = opp/adj
\(tan \theta = \frac{2}{\sqrt{21}}\)
