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Question

# If sin(α+β)=1,sin(α−β)=12, then tan(α+2β)tan(2α+β) is equal to α,βϵ(0,π/2)

A
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B
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C
0
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D
none of these
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Solution

## The correct option is B 1sin(α+β)=1⇒cos(α+β)=0⇒cos3(α+β)=0.........(1)sin(α+β)=12⇒cos(α−β)=√32.......(2)Now,tan(α+2β)+tan(2α+β)=sin(α+2β)sin(2α+β)cos(α+2β)cos(2α+β)=2sin(α+2β)sin(2α+β)2cos(α+2β)cos(2α+β) Using 2sinXsinY=cos(X−Y)−cos(X+Y)2cosXcosY=cos(X+Y)+cos(X−Y)=cos((α+2β)−(2α+β))−cos((α+2β)+(2α+β))cos((α+2β)+(2α+β))+cos((α+2β)−(2α+β))=cos(−α+β)−cos3(α+β)cos3(α+β)+cos(−α+β)=cos(α−β)−cos3(α+β)cos3(α+β)+cos(α−β)Substituting values from (1) and (2), we gettan(α+2β)tan(2α+β)=√32+00+√32=1

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