Question

Find the value of 4 $ta{n}^{-1}\frac{1}{5}-ta{n}^{-1}\frac{1}{239}.$

Answer 1

We have, $4ta{n}^{-1}\frac{1}{5}-ta{n}^{-1}\frac{1}{239}$

$=2.2ta{n}^{-1}\frac{1}{5}-ta{n}^{-1}\frac{1}{239}=2.\left[ta{n}^{-1}\frac{\frac{2}{5}}{1-{\left(\frac{1}{5}\right)}^2}\right]-ta{n}^{-1}\frac{1}{239}[\because 2ta{n}^{-1}x=ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)]=2.\left[ta{n}^{-1}\left(\frac{\frac{2}{5}}{1-\frac{1}{25}}\right)\right]-ta{n}^{-1}\frac{1}{239}=2.\left[ta{n}^{-1}\left(\frac{2/5}{24/25}\right)\right]-ta{n}^{-1}\frac{1}{239}=2ta{n}^{-1}\frac{5}{12}-ta{n}^{-1}\frac{1}{239}=ta{n}^{-1}\frac{2.\frac{5}{12}}{1-{\left(\frac{5}{12}\right)}^2}-ta{n}^{-1}\frac{1}{239}[\because 2ta{n}^{-1}x=ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)]=ta{n}^{-1}\left(\frac{\frac{5}{6}}{1-\frac{25}{144}}\right)-ta{n}^{-1}\frac{1}{239}=ta{n}^{-1}\left(\frac{144\times 5}{119\times 6}\right)-ta{n}^{-1}\frac{1}{239}=ta{n}^{-1}\left(\frac{120}{119}\right)-ta{n}^{-1}\frac{1}{239}$

$=ta{n}^{-1}\left(\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}.\frac{1}{239}}\right)[\because ta{n}^{-1}x-ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x-y}{1+xy}\right)]=ta{n}^{-1}\left(\frac{120\times 239-119}{119\times 239\times +120}\right)=ta{n}^{-1}\left[\frac{28680-119}{28441+120}\right]=ta{n}^{-1}\frac{28561}{28561}=ta{n}^{-1}\left(1\right)=ta{n}^{-1}\left(tan\frac{\pi }{4}\right)=\frac{\pi }{4}$