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?
 A: HINT:
$\cos A \cos B - \sin A \sin B=\cos(A+B)$
and $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
Do you know how to find $\cos \theta$ from $\tan\theta?$
A: It is given that tanA=1, and tan B=$/sqrt3$.
This implies that A can be 45 degrees and B can be 60 degrees.( Taking inverse, and using principal values)
Also cosA.cosB-sinAsinB=cos(A B)
So, cos(A B)=cos(60 45)=cos(105)=-sin(15)=-($\sqrt3$-1)/(2$\sqrt2$)
A: 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 $cos 60^0 = 1/2$.
A: Without some additional information, you quite simply can't. The closest you can come is using the identity $$\cos^2\theta=\frac1{\tan^2\theta+1}$$ (which is derived from the Pythagorean identity, and holds wherever $\tan\theta$ is defined), from which you can determine that $$\cos\theta=\pm\sqrt{\frac1{1+\tan^2\theta}}.$$ Now, if you have an additional assumption, such as that $\theta$ is acute, then this becomes $$\cos\theta=\sqrt{\frac1{1+\tan^2\theta}},$$ at which point you can use this together with the angle sum identity for cosine to solve your problem
