Question

${\displaystyle {\tan }^{-1} \frac{1}{4}+{\tan }^{-1} \frac{2}{9}=\frac{1}{2}{\cos }^{-1} \frac{3}{2}=\frac{1}{2}{\sin }^{-1}\left(\frac{4}{5}\right)}$

Answer 1

LHS = ${\displaystyle {\tan }^{-1} \frac{1}{4}+{\tan }^{-1} \frac{2}{9}}$

${\displaystyle ={\tan }^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}\times \frac{2}{9}}\right)}$     ${\displaystyle [\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)]}$

${\displaystyle ={\tan }^{-1}\left(\frac{\frac{17}{36}}{\frac{34}{36}}\right)}$

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

${\displaystyle =\frac{1}{2}{\cos }^{-1}\left(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}\right)}$         ${\displaystyle [\because {\tan }^{-1}x=\frac{1}{2}{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)]}$

${\displaystyle =\frac{1}{2}{\cos }^{-1}\left(\frac{\frac{3}{4}}{\frac{5}{4}}\right)}$

${\displaystyle =\frac{1}{2}{\cos }^{-1}\left(\frac{3}{5}\right)}$

Now,

${\displaystyle {\tan }^{-1} \frac{1}{2}=\frac{1}{2}{\sin }^{-1}\left(\frac{\frac{2}{2}}{1+\frac{1}{4}}\right)}$     ${\displaystyle [\because {\tan }^{-1}x=\frac{1}{2}{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)]}$

${\displaystyle =\frac{1}{2}{\sin }^{-1}\left(\frac{1}{\frac{5}{4}}\right)}$

${\displaystyle =\frac{1}{2}{\sin }^{-1}\left(\frac{4}{5}\right)}$