Question

Prove the following,

$2ta{n}^{-1}\left(\frac{1}{2}\right)+ta{n}^{-1}\left(\frac{1}{7}\right)=ta{n}^{-1}\left(\frac{31}{17}\right)$

Answer 1

Given $2ta{n}^{-1}\left(\frac{1}{2}\right)+ta{n}^{-1}\left(\frac{1}{7}\right)=ta{n}^{-1}\left(\frac{31}{17}\right)$
LHS = $2ta{n}^{-1}\left(\frac{1}{2}\right)+ta{n}^{-1}\left(\frac{1}{7}\right)=ta{n}^{-1}\left(\frac{2\times \frac{1}{2}}{1-{\left(\frac{1}{2}\right)}^2}\right)+ta{n}^{-1}\left(\frac{1}{7}\right)$

$(\because 2ta{n}^{-1}x=ta{n}^{-1}\left(\frac{2x}{1-x^2}\right))$

$=ta{n}^{-1}\frac{1}{1-\frac{1}{4}}+ta{n}^{-1}\frac{1}{7}=ta{n}^{-1}\left(\frac{4}{3}\right)+ta{n}^{-1}\left(\frac{1}{7}\right)$

$=ta{n}^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\times \frac{1}{7}}\right)(\because ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x+y}{1-xy}\right))$

$=ta{n}^{-1}\left(\frac{\frac{28+3}{21}}{1-\frac{4}{21}}\right)=ta{n}^{-1}\left(\frac{\frac{31}{21}}{\frac{17}{21}}\right)=ta{n}^{-1}(\frac{31}{21}\times \frac{21}{17})=ta{n}^{-1}\left(\frac{31}{17}\right)=RHS$
Hence proved