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Q.

If  sin3θ−cos3θsinθ−cosθ−cosθ1+cot2θ−2tanθcotθ=−1,θ∈[0,2π] then

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a

θ∈0,π2−π4

b

θ∈π2,π−3π4

c

θ∈π,3π2−5π4

d

θ∈(0,π)−π4,π2

answer is D.

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Detailed Solution

sinθ−cosθ≠0    ∴θ≠π4,5π4Now 1+cosθ(sinθ−|sinθ|)−2=−1cosθsinθ−|sinθ|=0∴θ∈(0,π)−π4,π2
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