Question

If A+B+C=π and cot2θ=cot2A+cot2B+cot2C, then sin2(A−θ)sin22A+sin2(B−θ)sin22B+sin2(C−θ)sin22C is equal tosin2θsecAsecBsecC2sin2θcosAcosBcosC2−sin2θcosAcosBcosC2−sin2θsecAsecBsecC2

Solution

The correct option is D −sin2θsecAsecBsecC2cot2θ=cot2A+cot2B+cot2C⇒cot2θ−cot2A=cot2B+cot2C ⇒cos2θsin2A−sin2θcos2Asin2θsin2A=sin2(B+C)sin2Bsin2C ⇒sin(2A−2θ)sin2θsin2A=sin(2π−2A)sin2Bsin2C[∵A+B+C=π] ⇒sin2(A−θ)sin22A=−sin2θsin2Bsin2C⋯(1) ​​​ Similarly ​​​​ sin2(B−θ)sin22B=−sin2θsin2Asin2C...(2) sin2(C−θ)sin22C=−sin2θsin2Asin2B...(3) Add (1),(2) and (3) sin(2A−2θ)sin22A+sin(2B−2θ)sin22B+sin(2C−2θ)sin22C =−sin2θsin2Bsin2C−sin2θsin2Asin2C−sin2θsin2Asin2B =−sin2θ[sin2A+sin2B+sin2Csin2Asin2Bsin2C] =−sin2θ[4sinAsinBsinC2sinAcosA⋅2sinBcosB⋅2sinCcosC] =−sin2θsecAsecBsecC2 Alternate solution: Assuming A=B=C=60∘, we get cot2θ=cot2A+cot2B+cot2C⇒cot2θ=cot120∘+cot120∘+cot120∘⇒cot2θ=3cot120∘⇒cot2θ=−3cot60∘⇒cot2θ=−3×1√3=−√3⇒cot2θ=cot150∘⇒2θ=150∘ sin(2A−2θ)sin22A+sin(2B−2θ)sin22B+sin(2C−2θ)sin22C=3×sin(120∘−150∘)sin2120∘=3×−sin30∘34=−2 From options, sin2θcosAcosBcosC=sin150∘(cos60∘)3=116 So, both the options with sin2θcosAcosBcosC is not coorect. Now, sin2θsecAsecBsecC=sin150∘(sec60∘)3=4 So,  sin(2A−2θ)sin22A+sin(2B−2θ)sin22B+sin(2C−2θ)sin22C=−sin2θsecAsecBsecC2

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