1. ## Calculate!

Given that $\frac{cos(A-B)}{cos(A+B)}=\frac{7}{5}$, prove that $cosAcosB = 6sinAsinB$ and deduce a relationship between $tanA$ and $tanB$. Given further that $A+B=45^\circ$, calculate the value of $tanA+tanB$.

I have already proved the equation.

I have also tried to deduce the relationship between tanA and tanB,

$cosAcosB=6sinAsinB$

$\frac{cosAcosB}{sinAsinB}=\frac{1}{6}$

[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 $tanA+tanB$

2. 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 $\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A\tan B}$.