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Question

If cos θtan θ=sec θ,then,θ is equal to


A

2nπ+3π2,nZ

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B

nπ+(1)nπ6,nZ

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C

nπ+π2,nZ

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D

none of these

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Solution

The correct option is B

nπ+(1)nπ6,nZ


Given equation:cos θtan θ=sec θcosθsinθsinθcosθ=1cosθcos2θsin2θsin θ cosθ=1cosθcos2θsin2θ=sinθ(1sin2θ)sin2θ=sinθ12sin2θ+sinθ1=02sin2θ+sinθ1=02sin2θ+2sinθsinθ1=02sinθ(sinθ+1)1(sinθ+1)=0(sinθ+1)(2sinθ1)=0sinθ+1=0 or 2sinθ1=0sinθ=1 or sinθ=12Now,sinθ=1sinθ=sin3π2θ=mπ+(1)m3π2,mZAnd,sin θ=12sinθ=sinπ6θ=nπ+(1)nπ6nZq=nπ+(1)nπ6,nZ


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