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Question

# If sin3θ−cos3θsinθ−cosθ−cosθ√1+cot2θ−2tanθcotθ=−1,θ∈[0,2π], then

A
θ(0,π2){π4}
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B
θ(π2,π){3π4}
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C
θ(π,3π2){5π4}
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D
θ(0,π){π4,π2}
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Solution

## The correct option is A θ∈(0,π)−{π4,π2}We know that sin3θ−cos3θ=(sinθ−cosθ)(sin2θ+cos2θ+sinθcosθ)=(sinθ−cosθ)(1+sinθcosθ)⇒sin3θ−cos3θsinθ−cosθ=1+sinθcosθBut for it to exist sinθ−cosθ≠0 , which implies θ≠π4The value of cosθ√1+cot2θ=|sinθ|cosθSo the given equation becomes 1+sinθcosθ−|sinθ|cosθ−2tanθcotθ=−1We know that tanθcotθ=1, remember that for θ=π2, tanθ does not exist.So the given equation reduces to sinθcosθ=|sinθ|cosθ⇒|sinθ|=sinθWhich implies 0<θ<π and θ≠π4,π2So the correct option is D

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