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Let Δ=1sinθ1sinθ1sinθ1sinθ1,0θ2π. The

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a
∆=0
b
Δ∈(2,∞)
c
Δ∈(2,4)
d
Δ∈[2,4]

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detailed solution

Correct option is D

Applying R1→R1+R3 to obtain Δ=002−sin⁡θ1sin⁡θ−1−sin⁡θ1=2sin2⁡θ+1As, 0≤sin2⁡θ≤1, we get Δ∈[2,4]


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