Let $$A($$θ$$)=$$$$sin$$⁡θicos⁡θicos⁡θ$$sin$$⁡θ, then

Correct option is 2A(θ)−1=A(π−θ)

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Let $A (θ) = (\begin{array}{cc} \sin θ & i \cos θ \\ i \cos θ & \sin θ \end{array}),$ then

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A(θ)−1=A(−θ)

A(θ)−1=A(π−θ)

A(θ)−1 does not exist

A(θ)2=A(2θ)

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detailed solution

Correct option is B

|A(θ)|=sin2⁡θ−i2cos2⁡θ=1∴ A(θ)−1=sin⁡θ−icos⁡θ−icos⁡θsin⁡θ =A(π−θ)

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Let A(θ)=sin⁡θicos⁡θicos⁡θsin⁡θ, then

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Let $A (θ) = (\begin{array}{cc} \sin θ & i \cos θ \\ i \cos θ & \sin θ \end{array}),$ then