Simplify:${tan}^{-1}(1/2)+{tan}^{-1}(1/3).$

We have,

${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}$

We know that,

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

So,

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

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

$={\tan }^{-1}\left(\frac{5}{5}\right)$

$={\tan }^{-1}1$

$={\tan }^{-1}\tan \frac{\pi }{4}$

$=\frac{\pi }{4}$