
Calculate!
Given that $\displaystyle \frac{cos(AB)}{cos(A+B)}=\frac{7}{5}$, prove that $\displaystyle cosAcosB = 6sinAsinB$ and deduce a relationship between $\displaystyle tanA$ and $\displaystyle tanB$. Given further that $\displaystyle A+B=45^\circ$, calculate the value of $\displaystyle tanA+tanB$.
I have already proved the equation.
I have also tried to deduce the relationship between tanA and tanB,
$\displaystyle cosAcosB=6sinAsinB$
$\displaystyle \frac{cosAcosB}{sinAsinB}=\frac{1}{6}$
[Math]tanAtanB=\frac{1}{6}[/tex]
But am not sure if I am on the right track. Lastly, need help with calculating the value of $\displaystyle tanA+tanB$

Yes, you are on the right track (except that the 6 should be 1/6 — remember that tan = sin/cos, not cos/sin). For the last part, use the formula $\displaystyle \tan(A+B) = \frac{\tan A + \tan B}{1\tan A\tan B}$.