If tan^-1 3 + tan^-1x = tan^-18, then x =

we have,

$tan^{-1} 3 + tan^{-1}x = tan^{-1}8$

$⇒tan^{-1} x = tan^{-1}8- tan^{-1}3$

$⇒tan^{-1} x = tan^{-1}\left(\frac{8-3}{1+24}\right) ⇒x =\frac{1}{5}$