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Question

If sin 2A=λ sin 2B, prove that tan (A+B)tan (A-B)=λ+1λ-1.

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Solution

Given:
sin 2A = λ sin 2B

sin2Asin2B=λ

sin2A+sin2Bsin2A-sin2B=λ+1λ-12sin2A+2B2cos2A-2B22sin2A-2B2cos2A+2B2= λ+1λ-1 sinA+sinB=2sinA+B2cosA-B2 and sinA-sinB=2sinA-B2cosA+B2sinA+BcosA-BsinA-BcosA+B=λ+1λ-1tanA+BcotA-B=λ+1λ-1

tanA+BtanA-B=λ+1λ-1

Hence proved.

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