If sin\(x\) = \(\frac{2}{5}\), then find the value of tan2\(x\)?

\(\frac{4\sqrt {21 }}{17}\)

tan2x = \(\frac{2tan\;x}{1-tan^2 x}\)

&  tan x = \(\frac{sin\;x}{cos\;x}\)

Now, sinx = \(\frac{2}{5}\),

cosx = \(\sqrt {1\;-\;sin^2 }\) = \(\frac{\sqrt{21}}{5}\)

tanx = \(\frac{2}{\sqrt{21}}\)

tan2x = \(\frac{2tanx}{1 - tan^2x}\)  = \(\frac{4\sqrt{21}}{17}\)