$\cos {100}^0. \cos {40}^0+ \sin {100}^0. \sin {40}^0$

$$\begin{gathered} \mathrm{w}.\mathrm{k}.\mathrm{t} \\ \cos \mathrm{A} \cos \mathrm{B} + \mathrm{sinA} \mathrm{Sin} \mathrm{B}= \cos (\mathrm{A}-\mathrm{B}) \\ \cos {100}^0\cos {40}^0+ \sin {100}^0 \sin {40}^0= \cos ({100}^0-{40}^0) \\ =\cos {60}^0 \\ =\frac{1}{2} \end{gathered}$$