Question

Prove that:
${\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9}=\frac{1}{2}{\cos }^{-1}\frac{3}{5}$

Answer 1

$LHS={\tan }^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}\times \frac{2}{9}}\right)$
$={\tan }^{-1}\left(\frac{\frac{17}{36}}{\frac{36-2}{36}}\right)={\tan }^{-1}(\frac{17}{36}\times \frac{36}{34})$
$={\tan }^{-1}\left(\frac{1}{2}\right)$
$=\frac{1}{2}.2{\tan }^{-1}\left(\frac{1}{2}\right)$
$=\frac{1}{2}{\cos }^{-1}\frac{1-{\left(\frac{1}{2}\right)}^2}{1+{\left(\frac{1}{2}\right)}^2}=\frac{1}{2}{\cos }^{-1}\frac{\frac{4-1}{4}}{\frac{4+1}{4}}$
$=\frac{1}{2}{\cos }^{-1}(\frac{3}{4}\times \frac{4}{5})$
$=\frac{1}{2}{\cos }^{-1}\left(\frac{3}{5}\right)=RHS$