sin 43° cos 47° + cos 43° sin 47° = ?

$\sin 43^{\circ} \cos 47^{\circ}+\cos 43^{\circ} \sin 47^{\circ}$

$=\sin \left(90^{\circ}-47^{\circ}\right) \cos 47^{\circ}+\cos \left(90^{\circ}-47^{\circ}\right) \sin 47^{\circ}$

$=\cos 47^{\circ} \cos 47^{\circ}+\sin 47^{\circ} \sin 47^{\circ} \quad\left(\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right.$ and $\left.\cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$

$=\cos ^{2} 47^{\circ}+\sin ^{2} 47^{\circ}$

$=1$    (using the identity: $\sin ^{2} \theta+\cos ^{2} \theta=1$ )

Hence, the correct option is (b).