Trigonometry question

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