TITLE: $\tan(a) = 3/4$ and $\tan (b) = 5/12$, what is $\cos(a+b)$
QUESTION [5 upvotes]: It is known that 
$$\tan(a) = \frac{3}{4}, \:\:\: \tan(b) = \frac{5}{12} $$
with $a,b < \frac{\pi}{2}$.
What is $\cos(a+b)$?

Attempt :
$$ \cos(a+b) = \cos(a) \cos(b) - \sin(a) \sin(b) $$
And we can write $\tan(a) = \sin(a)/\cos(a) = 0.3/0.4 $
and $ \sin(b)/\cos(b) = 0.05/0.12 $
so
$$ \cos(a+b) = (0.4)(0.12) - (0.3) (0.05)  = 33/1000$$
Is this correct? Thanks.

REPLY [3 votes]: $$ \tan(a+b)= \frac {\tan a +\tan b}{1-\tan a \tan b} = \frac {3/4 +5/12}{1-(3/4)(5/12)}= \frac {56}{33}$$
$$ \sec^2(a+b)=1+\tan^2(a+b)=\frac {4225}{1089}=(\frac {65}{33})^2$$
$$\cos(a+b)= 33/65$$