Question

tan-1 (1/4) +tan-1 (2/9) is equal to

Answer 1

make use of the identity,

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

Here $\tan A=\frac{1}{4}\\ \tan B =\frac{2}{9}$

$A+B=\tan ^{-1}(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4}\frac{2}{9}} )\\ A+B=\tan ^{-1}(\frac{17}{34} )\\ A+B=\tan ^{-1}(\frac{1}{2} )$

Thus,

$\tan ^{-1}(\frac{1}{4} )+\tan ^{-1}(\frac{2}{9} )=\tan ^{-1}(\frac{1}{2} )$