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Question Number 113200 by bobhans last updated on 11/Sep/20

$$\mathrm{prove}\mathrm{that}{2\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)=\frac{\pi }{4}$$

Answered by john santu last updated on 11/Sep/20

$$\left(Q\right)provethat2{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)=\frac{\pi }{4}.$$$$\left(sol\right)recall{\tan }^{-1}\left(\frac{1}{3}\right)=arg(3+i)$$$$\to 2{\tan }^{-1}\left(\frac{1}{3}\right)=2arg(3+i)=arg\left({(3+i)}^2\right)=arg(8+6i)$$$$=arg(4+3i)$$$$nowwehave2{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)=$$$$arg(4+3i)+arg(7+i)=$$$$arg\left((4+3i)(7+i)\right)=$$$$arg(25+25i)=arg(1+i)=\frac{\pi }{4}.\left(✓\right)$$

Answered by Dwaipayan Shikari last updated on 11/Sep/20

$$2{tan}^{-1}\frac{1}{3}={tan}^{-1}\left(\frac{\frac{1}{3}+\frac{1}{3}}{1-\frac{1}{9}}\right)={tan}^{-1}\frac{3}{4}$$$${tan}^{-1}\left(\frac{1}{7}\right)+{tan}^{-1}\left(\frac{3}{4}\right)={tan}^{-1}\left(\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{3}{28}}\right)={tan}^{-1}\left(1\right)=\frac{\pi }{4}$$

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