Question

Evaluate $\tan \left[2{\tan }^{-1}\left(\frac{1}{5}\right)-\left(\frac{\mathrm{\pi }}{4}\right)\right]$

• A. $\frac{17}{7}$
• B. $-\frac{17}{7}$
• C. $\frac{7}{17}$
• D. $-\frac{7}{17}$

Answer 1

The correct option is D

$-\frac{7}{17}$

Explanation for correct option

Simplifying the given expression

Given, $\tan \left[2{\tan }^{-1}\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)-\left(\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)\right]$

Equating the above expression, we get:

$$\begin{gathered} 2{\tan }^{-1}\left(\frac{1}{5}\right)={\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{5} \\ ={\tan }^{-1}\left[\frac{\frac{1}{5}+\frac{1}{5}}{1-\frac{1}{5}\times \frac{1}{5}}\right] \\ ={\tan }^{-1}\left[\frac{\frac{2}{5}}{1-\frac{1}{25}}\right] \\ ={\tan }^{-1}\left[\frac{2\times 25}{5}\right] \\ ={\tan }^{-1}\left(\frac{5}{12}\right) \end{gathered}$$

We know that,

$\tan (\mathrm{A}-\mathrm{B})=\frac{\mathrm{tanA}-\mathrm{tanB}}{1+\mathrm{tanAtanB}}$

Implementing the given expression in the above equation, we get,

$\tan \left[\left(2{\tan }^{-1}1/5\right)-\frac{\mathrm{\pi }}{4}\right]$$=\frac{\tan \left(2{\tan }^{-1}1/5\right)-\tan \frac{\mathrm{\pi }}{4}}{1+\tan \left(2{\tan }^{-1}\frac{1}{5}\right)\tan \frac{\mathrm{\pi }}{4}}$

$=\frac{\tan \left({\tan }^{-1}\frac{5}{12}\right)-1}{1+\tan \left({\tan }^{-1}\frac{5}{12}\right)\times 1}$

$=\frac{\left(\frac{5}{12}-1\right)}{\left(1+\frac{5}{12}\times 1\right)}$

$=\frac{\left(\frac{5-12}{12}\right)}{\left(\frac{12+5}{12}\right)}$

$=\frac{-7}{17}$

Hence, the correct answer is Option (D).