How to find $\cos A \cos B - \sin A \sin B$? Given that:
$\tan A=1$ 
and
$\tan B = \sqrt{3}$
How would you find $\cos A \cos B - \sin A \sin B$?

EDIT: This is what I've tried after reading bhattacharjee's answer:
$$ \tan(A+B) = \tan A+\tan B−\tan A\tan B$$
so,
  $\tan(A+B)= {1+\sqrt{3} \over 1-\sqrt{3}}$
from this I get $1 \over \cos(A+B)^2 $ $=1+ \left ( {1+\sqrt{3} \over 1-\sqrt{3}}\right )^2$
=> $ 1 \over \cos^2(A+B) $ $=$ $  8 \over 4-2 \sqrt{3}$ 
Is this right, because it seems like a dead end to me?  How am I supposed to proceed from here?

since $tan A =1$, then $A = 45^0$ and since $tan B = \sqrt{3}$, then $ B = 60^0$.
By drawing triangles you can find that $sin 45^0 = 1/\sqrt{2}$ and $cos 45^0 = 1/\sqrt{2}$ and $sin 60^0 = \sqrt{3}/2$ and $cos 60^0 = 1/2$.
