Question

$ta{n}^{-1}\frac{1}{2}+{tan}^{-1}\frac{1}{5}+{tan}^{-1}\frac{1}{8}=\frac{\pi }{4}$

Answer 1

${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{8}=\frac{\pi }{4}$

solving LHS

we know ${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{10}}\right)+{\tan }^{-1}\frac{1}{8}$

$\Rightarrow {\tan }^{-1}\left(\frac{7}{9}\right)+{\tan }^{-1}\frac{1}{8}={\tan }^{-1}\left(\frac{\frac{7}{9}+\frac{1}{8}}{1-\frac{7}{72}}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{65}{65}\right)=\frac{\pi }{4}=$RHS.