A) \[\pi \]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{4}\]
Correct Answer: D
Solution :
Since \[2{{\tan }^{-1}}x={{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}\] \ \[4{{\tan }^{-1}}\frac{1}{5}=2\,\left[ 2{{\tan }^{-1}}\frac{1}{5} \right]=2{{\tan }^{-1}}\frac{\frac{2}{5}}{1-\frac{1}{25}}\] \[=2{{\tan }^{-1}}\frac{10}{24}={{\tan }^{-1}}\frac{\frac{20}{24}}{1-\frac{100}{576}}={{\tan }^{-1}}\frac{120}{119}\] So, \[4{{\tan }^{-1}}\frac{1}{5}-{{\tan }^{-1}}\frac{1}{239}={{\tan }^{-1}}\frac{120}{119}-{{\tan }^{-1}}\frac{1}{239}\] \[={{\tan }^{-1}}\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}.\frac{1}{239}}={{\tan }^{-1}}\frac{(120\times 239)-119}{(119\times 239)+120}\] Þ \[{{\tan }^{-1}}1=\frac{\pi }{4}\].
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