Answer
$1$
Work Step by Step
The identity says: $\sin^2(\theta)+\cos^2(\theta)=1$.
We also know that $\cos(\theta)=\sin(90^{\circ}-\theta)$.
Hence,
$\cos^2(40^{\circ})+\cos^2(50^{\circ})
\\=\cos^2(40^{\circ})+\sin^2(90^{\circ}-50^{\circ})\\
=\cos^2(40^{\circ})+\sin^2(40^{\circ})
\\=1$
