unsure

Quote:

Originally Posted by mathAna1ys!5

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 $\tan a=M$ and $\tan b=N$ then,
$\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}=\frac{M+N}{1-MN}$ (1)

Now since $(3M+3)(2N+2)=12$ factor and divide thus, $(M+1)(N+1)=2$ now open parantheses (remember to FOIL) thus,
$MN+N+M+1=2$ thus,
$N+M=1-MN$ substitute that into (1):
thus,
$\tan (a+b)=\frac{1-MN}{1-MN}=1$
thus,
$a+b=\frac{\pi}{4}$ as one anwer.
Q.E.D.

(Note it was assumed that $a,b$ was non-coterminal was $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ ).