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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

If x ^ 3 θ+y ^ 3 θ= θ θ and x θ=y θ, then x^ 2 +y^ 2 =

If xsin⁡3θ+ycos⁡3θ=sin⁡θcos⁡θx \sin ^{3} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta and
xsin⁡θ=ycos⁡θ,x \sin \theta=y \cos \theta, then x2+y2=x^{2}+y^{2}=

Options

  1. A1
  2. B2
  3. C0
  4. DNone of these

Correct answer

A. 1

Step-by-step solution

xsin⁡3θ+ycos⁡3θ=sin⁡θcos⁡θx \sin ^{3} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta
and xsin⁡θ=ycos⁡θx \sin \theta=y \cos \theta
Equation (i) may be written as xsin⁡θ⋅sin⁡2θ+ycos⁡3θ=sin⁡θcos⁡θx \sin \theta \cdot \sin ^{2} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta
⇒ycos⁡θsin⁡2θ+ycos⁡3θ=sin⁡θcos⁡θ\Rightarrow y \cos \theta \sin ^{2} \theta+y \cos ^{3} \theta=\sin \theta \cos \theta
⇒ycos⁡θ(sin⁡2θ+cos⁡2θ)=sin⁡θcos⁡θ\Rightarrow y \cos \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\sin \theta \cos \theta
⇒ycos⁡θ=sin⁡θcos⁡θ∴y=sin⁡θ…\Rightarrow y \cos \theta=\sin \theta \cos \theta \therefore y=\sin \theta \quad \ldots
Putting the value of y from
(iii) in (ii), we get xsin⁡θ=sin⁡θ⋅cos⁡θ⇒x=cos⁡θ…\mathrm{x} \sin \theta=\sin \theta \cdot \cos \theta \Rightarrow \mathrm{x}=\cos \theta \quad \ldots (iv)
Squaring (iii) and (iv) and adding, we get \mathrmx2+\mathrmy2=cos⁡2θ+sin⁡2θ=1\mathrmx^{2}+\mathrmy^{2}=\cos ^{2} \theta+\sin ^{2} \theta=1
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