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Question Number 126499 by liberty last updated on 21/Dec/20

$$4\arctan \left(\frac{1}{5}\right)-\arctan \left(\frac{1}{239}\right)=?$$

Answered by benjo_mathlover last updated on 21/Dec/20

$$letx={\tan }^{-1}\left(\frac{1}{5}\right)\Rightarrow \tan x=\frac{1}{5}$$$$\tan 2x=\frac{2\tan x}{1-\tan {}^{2}x}=\frac{2/5}{1-1/25}$$$$\tan 2x=\frac{5}{12};\tan 4x=\frac{2\tan 2x}{1-\tan {}^2\left(2x\right)}$$$$\tan 4x=\frac{10/12}{1-25/144}=\frac{120}{119}$$$$\iff \frac{1}{239}=\tan (4x-\alpha )$$$$\iff \frac{1}{239}=\frac{\tan 4x-\tan \alpha }{1+\tan 4x.\tan \alpha }$$$$\iff \frac{1}{239}=\frac{\frac{120}{119}-\tan \alpha }{1+\frac{120}{119}\tan \alpha }$$$$\iff \frac{1}{239}=\frac{120-119\tan \alpha }{119+120\tan \alpha }$$$$weget\tan \alpha =1\wedge \alpha =\frac{\pi }{4}$$$$thus4{\tan }^{-1}\left(\frac{1}{5}\right)-{\tan }^{-1}\left(\frac{1}{239}\right)=\frac{\pi }{4}$$$$$$

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