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If θR, then maximum value of Δ=11111+sinθ1111+cosθ is

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a
3/2
b
1/2
c
1/2
d
None of these

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detailed solution

Correct option is B

Applying C2→C2−C1 and C3→C3−C1, we getΔ=1001sin⁡θ010cos⁡θ=sin⁡θcos⁡θ=(sin⁡2θ)/2So, maximum value of Δ is 1/2.


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