Δ=∣∣
∣
∣∣sin2AcotA1sin2BcotB1sin2CcotC1∣∣
∣
∣∣ApplyingR2→R2−R1andR3→R3−R1wegetΔ=∣∣
∣
∣
∣
∣∣sin2AcotA1sin(B+A)sin(B−A)sin(A−B)sinAsinB0sin(C+A)sin(C−A)sin(A−C)sinAsinC0∣∣
∣
∣
∣
∣∣[∵cotα−cotβ=sin(β−α)sinαsinβ]ExpandingalongC3,wegetΔ=sin(A−B)sin(A−C)sinA[−sin(B+A)sinC+sin(C+A)sinB]=sin(A−B)sin(A−C)sinA[−sin(π−C)sinC+sin(π−B)sinB]=sin(A−B)sin(A−C)sinA[−sinCsinC+sinBsinB]=0
⇒Δ+5=5