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Question

lf A+B+C=π, then∣ ∣ ∣sin2AcotA1sin2BcotB1sin2CcotC1∣ ∣ ∣

A
0
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B
1
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C
-1
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D
2
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Solution

The correct option is A 0
∣ ∣ ∣sin2AcotA1sin2BcotB1sin2CcotC1∣ ∣ ∣
R1R1R2,R2R2R3
=∣ ∣ ∣sin2Asin2BcotAcotB0sin2Bsin2CcotBcotC0sin2CcotC1∣ ∣ ∣
=∣ ∣ ∣ ∣ ∣sin(A+B)sin(AB)cosAsinBsinAcosBsinAsinB0sin(B+C)sin(BC)cosBsinCsinBcosCsinBsinC0sin2CcotC1∣ ∣ ∣ ∣ ∣
=∣ ∣ ∣ ∣ ∣sinCsin(AB)sin(AB)sinAsinB0sinAsin(BC)sin(BC)sinBsinC0sin2CcotC1∣ ∣ ∣ ∣ ∣
=sin(BC)sin(AB)sinB+sin(BC)sin(AB)sinB
=0

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