Answer
$\dfrac{\sqrt3}{2}$
Work Step by Step
Recall that $\cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)$.
Hence,
$\cos(40^\circ)\cos(10^\circ)+\sin(40^\circ)\sin(10^\circ)\\
=\cos(40^\circ-10^\circ)\\
=\cos(30^\circ)\\
=\dfrac{\sqrt3}{2}$
