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Question

If A=1sinθ1sinθ1sinθ1sinθ1 where 0θ2π then-

A
Det (A)=0
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B
Det (A)(0,)
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C
Det (A)[0,4]
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D
Det (A)[2,)
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Solution

The correct option is D Det (A)[0,4]
A=1sinθ1sinθ1sinθ1sinθ1

detA=1(1+sin2θ)sinθ(sinθsinθ)+1(sin2θ1)

=1+sin2θ+sin2θ+sin2θ+sin2θ1=4sin2θ

0sin2θ104sin2θ4

detA[0,4]

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