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Question: Answered & Verified by Expert
If Δ1=xsinθcosθ-sinθ-x1cosθ1x and Δ2=xsin2θcos2θ-sin2θ-x1cos2θ1x, x0; then for all θ0,π2 :
MathematicsDeterminantsJEE MainJEE Main 2019 (10 Apr Shift 1)
Options:
  • A Δ1+Δ2=-2(x3+x-1)
  • B Δ1-Δ2=x(cos2θ-cos4θ)
  • C Δ1+Δ2=-2x3
  • D Δ1-Δ2=-2x3
Solution:
1744 Upvotes Verified Answer
The correct answer is: Δ1+Δ2=-2x3
Δ1=xsinθ-sinθ-xcosθ1    cosθ1x
=x-x2-1-sinθ-xsinθ-cosθ+cosθ(-sinθ+xcosθ)
=-x3-x+xsin2θ+sinθcosθ-sinθcosθ+xcos2θ
=-x3-x+xsin2θ+cos2θ
=-x3
Similarly: Δ2=-x3
Δ1+Δ2=-2x3

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