${\cos }^{2}10°+{\mathrm{ cos}}^{2}50°-\cos 10°\cos 50°$

Using, $\cos 2x=2{\cos }^{2}x-1,$ $\Rightarrow {\cos }^{2}x=\frac{1+\cos 2x}{2},$ we get

$=\frac{1}{2}\left[1+\cos 20°+1+\cos 100°-2\cos 50°\cos 10°\right]$

$=\frac{1}{2}\left[2+\cos 100°+\cos 20°-\cos 60°-\cos 40°\right]$

$=\frac{1}{2}\left[2+\cos 100°+\cos 20°-\frac{1}{2}-\cos 40°\right]$

Using, $\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right),$ we get

$=\frac{1}{2}\left[\frac{3}{2}+2\cos {60}^{°}\cos 40°-\cos 40°\right]$

$=\frac{1}{2}\left[\frac{3}{2}+2\times \frac{1}{2}\times \cos 40°-\cos 40°\right]$

$=\frac{1}{2}\left[\frac{3}{2}+\cos 40°-\cos 40°\right]$

$=\frac{1}{2}\times \frac{3}{2}=\frac{3}{4}.$