A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{4}\]
Correct Answer: D
Solution :
[d] \[\therefore 4.ta{{n}^{-1}}\left( \frac{1}{5} \right)-{{\tan }^{-1}}\left( \frac{1}{239} \right)\] \[=2.2ta{{n}^{-1}}\left( \frac{1}{5} \right)-{{\tan }^{-1}}\left( \frac{1}{239} \right)\] \[=2{{\tan }^{-1}}\left( \frac{2\times \frac{1}{5}}{1-{{\left( \frac{1}{5} \right)}^{2}}} \right)-{{\tan }^{-1}}\left( \frac{1}{239} \right)\] \[=2{{\tan }^{-1}}\left( \frac{2}{5}\times \frac{25}{24} \right)-{{\tan }^{-1}}\left( \frac{1}{239} \right)=2.{{\tan }^{-1}}\left( \frac{5}{12} \right)-{{\tan }^{1}}\left( \frac{1}{239} \right)\]\[={{\tan }^{-1}}\left( \frac{\frac{2\times 5}{12}}{1-{{\left( \frac{5}{12} \right)}^{2}}} \right)-{{\tan }^{-}}\left( \frac{1}{239} \right)\] \[={{\tan }^{-1}}\left( \frac{10}{12}\times \frac{144}{119} \right)-{{\tan }^{-1}}\left( \frac{1}{139} \right)\] \[={{\tan }^{-1}}\left( \frac{120}{119} \right)-{{\tan }^{-1}}\left( \frac{1}{139} \right)\] \[={{\tan }^{-1}}\left( \frac{\frac{120}{190}-\frac{1}{239}}{1+\frac{120}{119}\times \frac{1}{139}} \right)={{\tan }^{-1}}\left( \frac{120\times 239-119}{239\times 119+120} \right)\]\[={{\tan }^{-1}}\left( \frac{28561}{28561} \right)\] \[=ta{{n}^{-1}}1=ta{{n}^{-1}}\left( tan\frac{\pi }{4} \right)=\frac{\pi }{4}\] Hence, option [d] is correct.
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