<s>[INST] $\tan(a) = 3/4$ 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.
 [/INST] No, this can't be correct. Remember that $\sin^2x+\cos^2x=1$ for all $x$; your values for the sine and cosine of $a$ and $b$ do not satisfy this relation.
It turns out that
$$\sin a=\frac35\qquad\cos a=\frac45$$
$$\sin b=\frac5{13}\qquad\cos b=\frac{12}{13}$$
and thus
$$\cos(a+b)=\cos a\cos b-\sin a\sin b=\frac45\cdot\frac{12}{13}-\frac35\cdot\frac5{13}=\frac{33}{65}$$
</s>