The number of solutions to the equation  tan−1x3+tan−1x2=tan−1xis

tan−1x3+tan−1x2=tan−1xOr tan−1x/3+x/21−x2/6=tan−1x⇒5x6−x2=x⇒x=0   or   x2−1=0⇒x=±1Therefore,  x={−1,0,1}⇒3  solutions