Let Δ=1sinθ1−sinθ1sinθ−1−sinθ1,0≤θ≤2π. The
∆=0
Δ∈(2,∞)
Δ∈(2,4)
Δ∈[2,4]
Applying R1→R1+R3 to obtain
Δ=002−sinθ1sinθ−1−sinθ1=2sin2θ+1
As, 0≤sin2θ≤1, we get Δ∈[2,4]