Question

The numerical value of tan $\left(2{\tan }^{-1}\frac{1}{5}-\frac{\pi }{4}\right)$ is

Answer 1

Solution :-

Let $A=tan\left[2ta{n}^{-1}\right(\frac{1}{5})-\frac{\pi }{4}]$

Let $2ta{n}^{-1}\left(\frac{1}{5}\right)=\theta$

$\therefore A=tan(\theta -\pi /4)=\frac{tan\theta -tan\pi /4}{1+tan\theta tan\pi /4}=\frac{tan\theta -1}{tan\theta +1}$

$\frac{\theta }{2}=\frac{2ta{n}^{-1}}{2}\left(\frac{1}{5}\right)=ta{n}^{-1}\left(\frac{1}{5}\right)=\frac{\theta }{2}\Rightarrow tan\frac{\theta }{2}=\frac{1}{5}$

But $tan\theta =\frac{2tan\theta /2}{1-ta{n}^2\theta /2}=\frac{2\cdot 1/5}{1-1/25}=\frac{2/5}{24/25}=\frac{5}{12}$

$\therefore A=\frac{tan\theta -1}{tan\theta +1}=\frac{5/12-1}{5/12+1}=\frac{-7}{17}$

$\therefore$ Numerical Value = $\left|\frac{-7}{17}\right|=7/17$