DinahParmar116
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# Find the exact value of tan (arcsin (two fifths))

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 $tan \theta = \frac{2}{\sqrt{21}}$