Question

Prove that tan inverse 1/2+tan inverse 1/5=1/2 cos inverse 16/65

Answer 1

consider,

$tan^{-1}\frac{1}{2}+tan^{-1}\frac{1}{5}$

Using,

$\boxed{\bf\,tan^{-1}x+tan^{-1}y=tan^{-1}(\frac{x+y}{1-xy})}}$

$=tan^{-1}(\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{2}*\frac{1}{5}})$

$=tan^{-1}(\frac{\frac{5+2}{10}}{1-\frac{1}{10}})$

$=tan^{-1}(\frac{\frac{7}{10}}{\frac{9}{10}})$

$=tan^{-1}(\frac{7}{9})=X(say)$

$\implies\,tanX=\frac{7}{9}$

Using

$\boxed{cos2A=\frac{1-tan^2A}{1+tan^2A}}$

$cos2X=\frac{1-tan^2X}{1+tan^2X}}$

$cos2X=\frac{1-\frac{49}{81}}{1+\frac{49}{81}}$

$cos2X=\frac{\frac{81-49}{81}}{\frac{81+49}{81}}$

$cos2X=\frac{\frac{32}{81}}{\frac{130}{81}}$

$cos2X=\frac{32}{130}$

$cos2X=\frac{16}{65}$

$\implies\,2X=cos^{-1}(\frac{16}{65})$

$\implies\,X=\frac{1}{2}cos^{-1}(\frac{16}{65})$

$\implies\boxed{\bf\,tan^{-1}\frac{1}{2}+tan^{-1}\frac{1}{5}=\frac{1}{2}cos^{-1}(\frac{16}{65})}$