I kind of think I know what I could do, but I am unsure how to proceed.
Q: Find a + b in radians if tan a = M and tan b = N and (3M+3)(2N+2)= 12.
If $\displaystyle \tan a=M$ and $\displaystyle \tan b=N$ then,Originally Posted by mathAna1ys!5
$\displaystyle \tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}=\frac{M+N}{1-MN}$ (1)
Now since $\displaystyle (3M+3)(2N+2)=12$ factor and divide thus, $\displaystyle (M+1)(N+1)=2$ now open parantheses (remember to FOIL) thus,
$\displaystyle MN+N+M+1=2$ thus,
$\displaystyle N+M=1-MN$ substitute that into (1):
thus,
$\displaystyle \tan (a+b)=\frac{1-MN}{1-MN}=1$
thus,
$\displaystyle a+b=\frac{\pi}{4}$ as one anwer.
Q.E.D.
(Note it was assumed that $\displaystyle a,b$ was non-coterminal was $\displaystyle \frac{\pi}{2}$ or $\displaystyle \frac{3\pi}{2}$).