$\tan(a) = 3/4$ and $\tan (b) = 5/12$, what is $\cos(a+b)$ 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.

Solve for sin(a), cos(a), sin(b) and cos(b) (like I said in comments):
$sin(a)=\frac{3}{5}$
$cos(a)=\frac{4}{5}$
$sin(b)=\frac{5}{13}$
$cos(b)=\frac{12}{13}$
Thus $$cos(a+b)=\frac{4}{5}\times\frac{12}{13}-\frac{3}{5}\times\frac{5}{13}=\frac{33}{65}$$
