1.

Write the value of\(tan(2tan^{-1}\frac{1}{5})\).

Answer»

Let tan θ = 1/5 

Given tan (2 tan-1 1/5) = tan 2θ 

We know that \(tan2\theta=\frac{2tan\theta}{1-tan^2\theta}\)

\(=\frac{2\times\frac{1}{5}}{1-\frac{1}{25}}\)

\(=\frac{\frac{2}{5}}{\frac{24}{25}}\)

\(=\therefore tan(2tan^{-1}\frac{1}{5})=\frac{5}{12}\)



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