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Question

tanα=asinβ1acosβ and tanβ=bsinα1bcosα then show that sinαsinβ=ab

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Solution

tanα=asinβ1acosβGiven


sinαcosα=asinβ1acosβ


sinαacosβsinα=asinβcosα


sinα=asin(β+α)(1)


Alsotanβ=bsinα1bcosα


sinβcosβ=bsinα1bcosα


sinβbcosαsinβ=bsinαcosβ


sinβ=bsin(α+β)(2)


Lhs=sinαsinβ=asin(α+β)bsin(α+β)[by (1)&(2)]


=ab=Rhs

Hence proof


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