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Question Number 116966 by bobhans last updated on 08/Oct/20

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

Answered by bemath last updated on 08/Oct/20

Answered by TANMAY PANACEA last updated on 08/Oct/20

$$2{tan}^{-1}\left(\frac{\frac{1}{5}+\frac{1}{5}}{1-\frac{1}{5}\times \frac{1}{5}}\right)-{tan}^{-1}\left(\frac{1}{239}\right)$$$$=2{tan}^{-1}\left(\frac{2\times 25}{5\times 24}\right)-{tan}^{-1}\left(\frac{1}{239}\right)$$$$={tan}^{-1}\left(\frac{\frac{5}{12}+\frac{5}{12}}{1-\frac{25}{144}}\right)-{tan}^{-1}\left(\frac{1}{239}\right)$$$$={tan}^{-1}\left(\frac{10}{12}\times \frac{144}{119}\right)-{tan}^{-1}\left(\frac{1}{239}\right)$$$$={tan}^{-1}\left(\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120\times 1}{119\times 239}}\right)={tan}^{-1}\left(\frac{120\times 239-119}{119\times 239+120}\right)$$$$={tan}^{-1}\left(\frac{119\times 239+239-119}{119\times 239+120}\right)={tan}^{-1}\left(\frac{119\times 239+120}{119\times 239+120}\right)$$$$={tan}^{-1}\left(1\right)=\frac{\pi }{4}$$

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