If cosθ=cosα+cosβ1+cosαcosβ
then prove that one of the values of tanθ2 is tanα2tanβ2.
I don't even know how to start this question. Pls help
Answer
I think i have found out how to approach it. I hope this proof is satifactory.
Using the half angle formula,
tanθ2=±√1−cosθ1+cosθ⟶eq.1
Evaluating 1−cosθ1+cosθ first,
1−cosθ1+cosθ=1−(cosα+cosβ1+cosαcosβ)1+(cosα+cosβ1+cosαcosβ) [since cosθ=cosα+cosβ1+cosαcosβ]
=1+cosαcosβ−cosα−cosβ1+cosαcosβ+cosα+cosβ
=(1−cosα)(1−cosβ)(1+cosα)(1+cosβ)
Substituting this value of 1−cosθ1+cosθ into equation 1,
tanθ2=√(1−cosα)(1−cosβ)(1+cosα)(1+cosβ)
=√(1−cosα)2(1−cosβ)2(1−cos2α)(1−cos2β)
=±(1−cosα)(1−cosβ)sinαsinβ
Taking the positive value of tanθ2,
(1−cosα)(1−cosβ)sinαsinβ=4sin2α2sin2β24sinα2cosα2sinβ2cosβ2(using half angle and double angle formula)
=tanα2tanβ2
Therefore, tanα2tanβ2 is one of the values of tanθ2.
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