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Question

If A+B+C=π and cot2θ=cot2A+cot2B+cot2C, then sin2(Aθ)sin22A+sin2(Bθ)sin22B+sin2(Cθ)sin22C is equal to
  1. sin2θsecAsecBsecC2
  2. sin2θcosAcosBcosC2
  3. sin2θcosAcosBcosC2
  4. sin2θsecAsecBsecC2


Solution

The correct option is D sin2θsecAsecBsecC2
cot2θ=cot2A+cot2B+cot2Ccot2θcot2A=cot2B+cot2C

cos2θsin2Asin2θcos2Asin2θsin2A=sin2(B+C)sin2Bsin2C

sin(2A2θ)sin2θsin2A=sin(2π2A)sin2Bsin2C[A+B+C=π]

sin2(Aθ)sin22A=sin2θsin2Bsin2C(1)
​​​
Similarly
​​​​
sin2(Bθ)sin22B=sin2θsin2Asin2C...(2)

sin2(Cθ)sin22C=sin2θsin2Asin2B...(3)

Add (1),(2) and (3)

sin(2A2θ)sin22A+sin(2B2θ)sin22B+sin(2C2θ)sin22C

=sin2θsin2Bsin2Csin2θsin2Asin2Csin2θsin2Asin2B

=sin2θ[sin2A+sin2B+sin2Csin2Asin2Bsin2C]

=sin2θ[4sinAsinBsinC2sinAcosA2sinBcosB2sinCcosC]

=sin2θsecAsecBsecC2


Alternate solution:
Assuming A=B=C=60, we get
cot2θ=cot2A+cot2B+cot2Ccot2θ=cot120+cot120+cot120cot2θ=3cot120cot2θ=3cot60cot2θ=3×13=3cot2θ=cot1502θ=150

sin(2A2θ)sin22A+sin(2B2θ)sin22B+sin(2C2θ)sin22C=3×sin(120150)sin2120=3×sin3034=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(2A2θ)sin22A+sin(2B2θ)sin22B+sin(2C2θ)sin22C=sin2θsecAsecBsecC2

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