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Question Number 111725 by Aina Samuel Temidayo last updated on 04/Sep/20

$$\mathrm{Find}\mathrm{the}\mathrm{positive}\mathrm{integer}\mathrm{n}\mathrm{such}\mathrm{that}$$$${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{4}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{\mathrm{n}}\right)=\frac{\pi }{4}$$

Answered by $@y@m last updated on 04/Sep/20

$$Let{\tan }^{-1}\left(\frac{1}{3}\right)=\alpha \Rightarrow \tan \alpha =\frac{1}{3}$$$$Let{\tan }^{-1}\left(\frac{1}{4}\right)=\beta \Rightarrow \tan \beta =\frac{1}{4}$$$$Let{\tan }^{-1}\left(\frac{1}{5}\right)=\gamma \Rightarrow \tan \gamma =\frac{1}{5}$$$$Let{\tan }^{-1}\left(\frac{1}{n}\right)=\delta \Rightarrow \tan \delta =\frac{1}{n}$$$$ATQ,$$$$\alpha +\beta +\gamma +\delta =\frac{\pi }{4}$$$$\alpha +\beta +\gamma =\frac{\pi }{4}-\delta$$$$\tan (\alpha +\beta +\gamma )=\tan (\frac{\pi }{4}-\delta )$$$$\frac{\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{3.4.5}}{1-\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.3}}=\frac{1-\tan \delta }{1+\tan \delta }$$$$\frac{1-\tan \delta }{1+\tan \delta }=\frac{\frac{20+15+12-1}{60}}{\frac{60-5-3-4}{60}}=\frac{46}{48}=\frac{23}{24}$$$$24(1-\tan \delta )=23(1+\tan \delta )$$$$1=47\tan \delta$$$$\tan \delta =\frac{1}{47}$$$$\delta ={\tan }^{-1}\left(\frac{1}{47}\right)$$$$n=47$$

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

$$47\mathrm{is}\mathrm{part}\mathrm{of}\mathrm{the}\mathrm{options}\mathrm{I}\mathrm{have}\mathrm{here}.$$$$\mathrm{Thanks}.$$

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