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Question Number 34646 by JOHNMASANJA last updated on 09/May/18

$$showthat2{tan}^{-1}2+{tan}^{-1}3=$$$$\mathrm{\Pi }+{tan}^{-1}\frac{1}{3}$$

Answered by ajfour last updated on 09/May/18

$$let2\theta +\varphi ={2\tan }^{-1}2+{\tan }^{-1}3$$$$\Rightarrow \tan (2\theta +\varphi )=\frac{\tan 2\theta +\tan \varphi }{1-\tan 2\theta \tan \varphi }$$$$=\frac{\left(\frac{2\tan \theta }{1-\tan {}^2\theta }\right)+\tan \varphi }{1-\left(\frac{2\tan \theta }{1-\tan {}^2\theta }\right)\tan \varphi }$$$$=\frac{\frac{4}{1-4}+3}{1-\left(\frac{4}{1-4}\right)3}=\frac{5}{15}=\frac{1}{3}$$$$2\theta +\varphi ={2\tan }^{-1}2+{\tan }^{-1}3=\pi +{\tan }^{-1}\frac{1}{3}.$$

Commented by JOHNMASANJA last updated on 09/May/18

$$ididunderstandhowdoes\pi came$$$$from$$

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