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?
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Now apply with and . You will also eventually need .
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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