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Question

If cotθtanθ=secθ, then θ =

A
nπ+(1)nπ/6
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B
nπ+π/2
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C
2nπ+3π/2
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D
None of these
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Solution

The correct option is A nπ+(1)nπ/6
cotθtanθ=secθ
cosθsinθsinθcosθ=1cosθ
cos2θsin2θsinθcosθ=1cosθ
cosθ(cos2θsin2θ)sinθcosθ=0
cosθ=0 or cos2θsin2θsinθcosθ=0
cos2θsin2θ=sinθ
12sin2θ=sinθ
2sin2θ+sinθ1=0
sinθ=1±1+84=1±94
sinθ=1±34=12 or 1
As θπ2 so,
sinθ=12
θ=π6+nπ for 1st and 2nd Quadrant
θ=nπ+(1)nπ6

1154852_1182396_ans_b3285a76ab6f49f5b2621ddd6576f065.jpeg

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