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Q.

The maximum value of Δ = 1                  1           1 1            1 + sinθ     11+cosθ      1           1  is (where , θ is real number)

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Detailed Solution

Δ = 1                  1           1 1            1 + sinθ     11+cosθ      1           1 = 0            0           1 0           sin θ       1cos θ      0           1                                                     ∵C1→C1−C3 and C2→C2-C3=-(sinθ·cosθ)=-12·2sinθcosθ=-12sin2θSince the maximum value of sin2θis 1. So far maximum value of θ   should be -45°.Δ=-12sin2.-45°=-12sin-90°=12·1=12
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