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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2019 · Mathematics · Trigonometric Ratios & Identities

Paragraph: If p=X θ-Y θ, q=X θ+Y θ and p^2+4 pq+q^2=A X^2+B Y^2, 0...

Paragraph: If p=Xcos⁡θ−Ysin⁡θ,q=Xsin⁡θ+Ycos⁡θ\mathrm{p}=\mathrm{X} \cos \theta-\mathrm{Y} \sin \theta, \mathrm{q}=X \sin \theta+\mathrm{Y} \cos \theta and p2+4pq+q2=AX2+BY2,0≤0≤π2\mathrm{p}^2+4 \mathrm{pq}+\mathrm{q}^2=A X^2+B Y^2, 0 \leq 0 \leq \frac{\pi}{2}
Question:What is the value of A?

Options

  1. A4
  2. B3
  3. C2
  4. D1

Correct answer

B. 3

Step-by-step solution

p=xcos⁡θ−ysin⁡θq=xsin⁡θ+ycos⁡θGiven, p2+4pq+q2=Ax2+By2\begin{aligned} & \mathrm{p}=\mathrm{x} \cos \theta-\mathrm{y} \sin \theta \\ & \mathrm{q}=\mathrm{x} \sin \theta+\mathrm{y} \cos \theta \\ & \text{Given, } \mathrm{p}^2+4 p q+\mathrm{q}^2=A x^2+B y^2\end{aligned} Let us take θ=π4\theta=\frac{\pi}{4}. p=xcos⁡π4−ycos⁡π4=x−y2q=xsin⁡π4+ycos⁡π4=x+y2pq=x2−y22⇒2pq=x2−y2\begin{aligned} & p=x \cos \frac{\pi}{4}-y \cos \frac{\pi}{4}=\frac{x-y}{\sqrt{2}} \\ & q=x \sin \frac{\pi}{4}+y \cos \frac{\pi}{4}=\frac{x+y}{\sqrt{2}} \\ & p q=\frac{x^2-y^2}{2} \Rightarrow 2 p q=x^2-y^2\end{aligned}\Rightarrow 4 \mathrm{pq}=2 \mathrm{x}^2-2 \mathrm{y}^2 ...(1) Now, p2+q2=x2cos⁡2θ+y2sin⁡2θ−2xycos⁡θsin⁡θ+x2sin⁡2θ+y2cos⁡2θ+2xysin⁡θcos⁡θ=x2+y2...(2)Fromp^2+q^2=x^2 \cos ^2 \theta+y^2 \sin ^2 \theta-2 x y \cos \theta \sin \theta+x^2 \sin ^2 \theta+y^2 \cos ^2 \theta+2 x y \sin \theta \cos \theta=x^2+y^2...(2) From(1), (2), p2+q2+4pq=x2+y2+2x2−2y2=3x2−y2\begin{aligned} p^2+q^2+4 p q & =x^2+y^2+2 x^2-2 y^2 \\ & =3 x^2-y^2\end{aligned} Comparing this with the given form, we getθ=\theta=\frac{\pi}{4}$, A=3, B=-1
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