If cosθ=cosα−cosβcosαcosβ, then prove that tan(θ/2)=±tan(α/2)cot(β/2)
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Solution
tan2θ2=1−cosθ1+cosθ=(1−cosα−cosβ1−cosαcosβ)/(1+cosα−cosβ1−cosαcosβ) =(1−cosα)(1+cosβ)(1+cosα)(1−cosβ) or tan2θ2=2sin2(α/2)2cos2(α/2).2cos2(β/2)2sin2(β/2)=tan2(α/2)=±tan(α/2)cot(β/2)