Given tan(A-B)=2 and tan(A-C)=3. Find tan(3A-2B-C).
I have no clues how to solve.. can anyone help me?
$\displaystyle \displaystyle \tan{(3A - 2B - C)} = \tan{(2A - 2B + A - C)} = \tan{[2(A-B) + (A - C)]}$
Now apply $\displaystyle \displaystyle \tan{(\alpha \pm \beta)} \equiv \frac{\tan{\alpha} \pm \tan{\beta}}{1 \mp \tan{\alpha}\tan{\beta}}$ with $\displaystyle \displaystyle \alpha = 2(A - B)$ and $\displaystyle \displaystyle \beta = A - C$. You will also eventually need $\displaystyle \displaystyle \tan{2\theta} \equiv \frac{2\tan{\theta}}{1 - \tan^2{\theta}}$.