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Question

If A+B+C=π and
Δ=∣ ∣ ∣sin2AcotA1sin2BcotB1sin2CcotC1∣ ∣ ∣
find Δ+5

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Solution

Δ=∣ ∣ ∣sin2AcotA1sin2BcotB1sin2CcotC1∣ ∣ ∣ApplyingR2R2R1andR3R3R1wegetΔ=∣ ∣ ∣ ∣ ∣sin2AcotA1sin(B+A)sin(BA)sin(AB)sinAsinB0sin(C+A)sin(CA)sin(AC)sinAsinC0∣ ∣ ∣ ∣ ∣[cotαcotβ=sin(βα)sinαsinβ]ExpandingalongC3,wegetΔ=sin(AB)sin(AC)sinA[sin(B+A)sinC+sin(C+A)sinB]=sin(AB)sin(AC)sinA[sin(πC)sinC+sin(πB)sinB]=sin(AB)sin(AC)sinA[sinCsinC+sinBsinB]=0
Δ+5=5

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