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Question Number 49751 by Pk1167156@gmail.com last updated on 10/Dec/18

$$\mathrm{The}\mathrm{value}\mathrm{of}{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}\mathrm{is}$$

Commented by maxmathsup by imad last updated on 10/Dec/18

$$wehavetan(arctan\left(\frac{1}{2}\right)+arctan\left(\frac{1}{3}\right))=$$$$=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1\Rightarrow arctan\left(\frac{1}{2}\right)+arctan\left(\frac{1}{3}\right)=arctan\left(1\right)=\frac{\pi }{4}$$$$$$

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Dec/18

$${tan}^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times \frac{1}{3}}\right)={tan}^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right)={tan}^{-1}\left(1\right)=\frac{\pi }{4}$$

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