# Math Help - Trigonometry - compound angels

1. ## Trigonometry - compound angels

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?

2. $\displaystyle \tan{(3A - 2B - C)} = \tan{(2A - 2B + A - C)} = \tan{[2(A-B) + (A - C)]}$

Now apply $\displaystyle \tan{(\alpha \pm \beta)} \equiv \frac{\tan{\alpha} \pm \tan{\beta}}{1 \mp \tan{\alpha}\tan{\beta}}$ with $\displaystyle \alpha = 2(A - B)$ and $\displaystyle \beta = A - C$. You will also eventually need $\displaystyle \tan{2\theta} \equiv \frac{2\tan{\theta}}{1 - \tan^2{\theta}}$.

3. A - B = Tan' 2 and A - C = Tan' 3 (Tan' means tangent inverse)

3A - 2B - C = 2( A - B ) + ( A - C) = 2Tan'(2) + Tan'(3)

Now using identities 2Tan'A = Tan' (2A/1-A^2) and Tan'(A) + Tan'(B) = Tan' [(A + B)/(1 - AB)]