Question

tan-1(1/7)+2tan-1(1/3)=??​

Answer 1

We have the formula,

• $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\dfrac{x+y}{1-xy}\right)\quad\quad\dots(i)$

For $x=y,$

• $2\tan^{-1}x=\tan^{-1}\left(\dfrac{2x}{1-x^2}\right)\quad\quad\dots(ii)$

Given to find,

$\longrightarrow\tan^{-1}\left(\dfrac{1}{7}\right)+2\tan^{-1}\left(\dfrac{1}{3}\right)=\,?$

Using $(ii),$

$\longrightarrow\tan^{-1}\left(\dfrac{1}{7}\right)+2\tan^{-1}\left(\dfrac{1}{3}\right)=\tan^{-1}\left(\dfrac{1}{7}\right)+\tan^{-1}\left(\dfrac{2\times\frac{1}{3}}{1-\left(\frac{1}{3}\right)^2}\right)$

$\longrightarrow\tan^{-1}\left(\dfrac{1}{7}\right)+2\tan^{-1}\left(\dfrac{1}{3}\right)=\tan^{-1}\left(\dfrac{1}{7}\right)+\tan^{-1}\left(\dfrac{\frac{2}{3}}{1-\frac{1}{9}}\right)$

$\longrightarrow\tan^{-1}\left(\dfrac{1}{7}\right)+2\tan^{-1}\left(\dfrac{1}{3}\right)=\tan^{-1}\left(\dfrac{1}{7}\right)+\tan^{-1}\left(\dfrac{3}{4}\right)$

Using $(i),$

$\longrightarrow\tan^{-1}\left(\dfrac{1}{7}\right)+2\tan^{-1}\left(\dfrac{1}{3}\right)=\tan^{-1}\left(\dfrac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\times\frac{3}{4}}\right)$

$\longrightarrow\tan^{-1}\left(\dfrac{1}{7}\right)+2\tan^{-1}\left(\dfrac{1}{3}\right)=\tan^{-1}\left(\dfrac{\frac{25}{28}}{1-\frac{3}{28}}\right)$

$\longrightarrow\tan^{-1}\left(\dfrac{1}{7}\right)+2\tan^{-1}\left(\dfrac{1}{3}\right)=\tan^{-1}(1)$

$\longrightarrow\underline{\underline{\tan^{-1}\left(\dfrac{1}{7}\right)+2\tan^{-1}\left(\dfrac{1}{3}\right)=\dfrac{\pi}{4}}}$

Hence $\dfrac{\pi}{4}$ is the answer.