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Question Number 48527 by Pk1167156@gmail.com last updated on 25/Nov/18

$$\mathrm{If}xy+yz+zx=1,\mathrm{then}$$$${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=$$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18

$${tan}^{-1}x+{tan}^{-1}\left(\frac{y+z}{1-yz}\right)$$$$={tan}^{-1}\left(\frac{x+\frac{y+z}{1-yz}}{1-x\times \frac{y+z}{1-yz}}\right)$$$$={tan}^{-1}\left(\frac{x-xyz+y+z}{1-yz-xy-xz}\right)$$$${tan}^{-1}\left(\frac{x+y+z-xyz}{0}\right)$$$$={tan}^{-1}\left(\mathrm{\infty }\right)$$$$={tan}^{-1}\left(tan\frac{\pi }{2}\right)$$$$=\frac{\pi }{2}$$

Answered by ajfour last updated on 25/Nov/18

$${\tan }^{-1}x=\theta ,{\tan }^{-1}y=\varphi ,{\tan }^{-1}z=\psi$$$$let\delta =\theta +\varphi +\psi$$$$\tan \delta =\tan (\overline{\theta +\varphi }+\psi )$$$$=\frac{\tan (\theta +\varphi )+\tan \psi }{1-\tan (\theta +\varphi )\tan \psi }$$$$=\frac{\frac{\tan \theta +\tan \varphi }{1-\tan \theta \tan \varphi }+\tan \psi }{1-\tan (\theta +\varphi )\tan \psi }$$$$=\frac{\mathrm{\Sigma }\tan \theta -\mathrm{\Pi }\tan \theta }{1-\mathrm{\Sigma }\tan \theta \tan \varphi }$$$$but\mathrm{\Sigma }\tan \theta \tan \varphi =\mathrm{\Sigma }xy=1$$$$\Rightarrow \mathit{\delta }=\theta +\varphi +\psi =\pm \frac{\pi }{2}.$$

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