If a, b and c are the sides of a triangle and A, B and C are the angles opposite to a, b and c respectively, then Δ=∣∣
∣∣a2bsinAcsinAbsinA1cosAcsinAcosA1∣∣
∣∣ is independent of
A
a
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B
b
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C
c
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D
A, B, C
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Solution
The correct options are A a B b C c D A, B, C Δ=∣∣
∣∣a2bsinAcsinAbsinA1cosAcsinAcosA1∣∣
∣∣ Using Sin law sinA=ak,sinB=bk and sinC=ck Δ=∣∣
∣∣a2bakcakbak1cosAcakcosA1∣∣
∣∣=a2∣∣
∣∣1bkckbk1cosAckcosA1∣∣
∣∣=a2∣∣
∣∣1sinBsinCsinB1cosAsinCcosA1∣∣
∣∣ Applying C2→C2−sinBC1,C3→C3−sinCC1 Δ=a2∣∣
∣
∣∣100sinB1−sin2BcosA−sinBsinCsinCcosA−sinBsinC1−sin2C∣∣
∣
∣∣Δ=a2(cos2Bcos2C−(cosA−sinBsinC)2) Now cosA=cos(π−(B+C))=−cos(B+C)=−(cosBcosC−sinBsinC)⇒cosA−sinBsinC=−cosBcosC Substituting this we get Δ=0