Question

Show that $ta{n}^{-1}\frac{1}{2}+ta{n}^{-1}\frac{2}{11}+ta{n}^{-1}\frac{4}{3}=\frac{\pi }{2}$.

Answer 1

${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-AB}\right)$

$\Rightarrow {\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{2}$

$={\tan }^{-1}\left(\frac{\frac{4}{3}+\frac{1}{2}}{1-\frac{4}{3}\times \frac{1}{2}}\right)={\tan }^{-1}\left(\frac{\frac{11}{6}}{1-\frac{2}{3}}\right)$

$={\tan }^{-1}\frac{11}{2}$

${\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{2}{11}={\tan }^{-1}\frac{11}{2}+{\tan }^{-1}\frac{2}{11}$

$={\tan }^{-1}\left(\frac{\frac{11}{2}+\frac{2}{11}}{1-\frac{11}{2}\times \frac{2}{11}}\right)={\tan }^{-1}\infty =\frac{\pi }{2}$