Trigonometry - compound angels

**MichaelLight** · June 7th, 2011, 09:50 PM

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?

**Prove It** · June 7th, 2011, 10:04 PM

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

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 $\displaystyle \tan{2\theta} \equiv \frac{2\tan{\theta}}{1 - \tan^2{\theta}}$ .

**waqarhaider** · June 7th, 2011, 11:58 PM

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)]

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