F.4 trigonometry===1

2010-03-08 12:09 am
Let tanA = sinα / (α-cosα) and tanB = asinα / (1-acosα), where a>1 and cosα ≠ 1/a. Prove that tan(A+B) = -tanα

回答 (1)

2010-03-08 1:35 am
✔ 最佳答案
tanA + tanB
= sinα / (a-cosα) + asinα / (1-acosα)
= [(sinα)(1-acosα) + (asinα)(a-cosα)] / [(a-cosα)(1-acosα)]
= (sinα)(1-acosα + a(a - cosα)) / [(a-cosα)(1-acosα)]
= (sinα)(1 + a^2 - 2acosα) / [(a-cosα)(1-acosα)]

1 - tanA tanB
= 1 - [sinα / (a-cosα)] [asinα / (1-acosα)]
= 1 - [a (sinα)^2] / [(a-cosα)(1-acosα)]
= [a - cosα - (cosα)a^2 + a(cosα)^2 - a (sinα)^2] / [(a-cosα)(1-acosα)]
= [a - cosα - (cosα)a^2 + a(cosα)^2 - a(1 - (cosα)^2)] / [(a-cosα)(1-acosα)]
= [- cosα - (cosα)a^2 + 2a(cosα)^2 ] / [(a-cosα)(1-acosα)]
= (- cosα)(1 + a^2 - 2acosα) / [(a-cosα)(1-acosα)]
So
tan(A+B) = (tanA + tanB) / (1 - tanA tanB)
= (sinα)(1 + a^2 - 2acosα) / (- cosα)(1 + a^2 - 2acosα)
= - sinα / cosα
= - tan α


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