Answer
$\sin(A+B)=\sin A+\sin B$
is not an identity.
Work Step by Step
$\sin(30^{o}+60^{o})=\sin 90^{o}=1$
$\displaystyle \sin(30^{o})+\sin(60^{o})=\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{1+\sqrt{3}}{2}\neq 1$
So,
$\sin(A+B)=\sin A+\sin B$
is not an identity.
($A$ counterexample is $A=30^{o}, B=60^{o})$