Ans:
Let $P(x_1,y_1)$ be a point in first quadrant which is at a distance of 1 unit from origin O
and let angle made by OP with X axis be A.
Similarly let $Q(x_2,y_2)$ be a point in $1^{st}$ quadrant which is at a distance of 1 unit from origin O
and let angle made by OQ with X axis be B.
$\Rightarrow |\overrightarrow {OP}|=|\overrightarrow {OQ}|=1$ and $ <\:POQ=A-B$
$\Rightarrow \overrightarrow {OP}=cosA\hat i+sinA\hat j$ and $\overrightarrow {OQ}=cosB\hat i+sinB\hat j$.
We know that $\overrightarrow a.\overrightarrow b=|\overrightarrow a||\overrightarrow b|cos\theta$ where $\theta$ is the angle between the vectors.
$\Rightarrow \overrightarrow {OP}\times\overrightarrow {OQ}=|\overrightarrow {OP}||\overrightarrow {OQ}|cos(A-B)$
$\Rightarrow\:\overrightarrow {OP}\times\overrightarrow {OQ}=cos(A-B)$...............(i)
We also know that $\overrightarrow a.\overrightarrow b=a_1b_1+a_2b_2+a_3b_3$
$\Rightarrow\:\overrightarrow {OP}\times\overrightarrow {OQ} =(cosA\hat i+sinA\hat j).(cosB\hat i+sinB\hat j)$
$=cosAcosB+sinAcosB$
$\Rightarrow \:\overrightarrow {OP}\times\overrightarrow {OQ}=cosAcosB+sinAsinB$................(ii)
From (i) and (ii) we get the result
$cos(A-B)=cosAcosB+sinAsinB$.
Hence proved.