10 \(\cos \left ( A\right ) \cos \left ( B\right ) \)

Adding (3)+(3A) gives

\begin{align*} \cos \left ( A+B\right ) & =\cos A\cos B-\sin A\sin B\\ \cos \left ( A-B\right ) & =\cos A\cos B+\sin A\sin B\\ \cos \left ( A+B\right ) +\cos \left ( A-B\right ) & =2\cos A\cos B \end{align*}

Hence

\[ \cos A\cos B=\frac {1}{2}\left ( \cos \left ( A+B\right ) +\cos \left ( A-B\right ) \right ) \]