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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2020 · Mathematics · Trigonometric Ratios & Identities

Paragraph: Read the following information and answer that follow:...

Paragraph: Read the following information and answer that follow: Let asin⁡2x+bcos⁡2x=c;bsin⁡2y+acos⁡2y=da \sin ^2 x+b \cos ^2 x=c ; b \sin ^2 y+a \cos ^2 y=d and ptan⁡x=qtan⁡yp \tan x=q \tan y.
Question: What is p2q2\frac{p^2}{q^2} equal to?

Options

  1. A(b−c)(b−d)(a−d)(a−c)\frac{(b-c)(b-d)}{(a-d)(a-c)}
  2. B(a−d)(c−a)(b−c)(d−b)\frac{(a-d)(c-a)}{(b-c)(d-b)}
  3. C(d−a)(c−a)(b−c)(d−b)\frac{(d-a)(c-a)}{(b-c)(d-b)}
  4. D(b−c)(b−d)(c−a)(a−d)\frac{(b-c)(b-d)}{(c-a)(a-d)}

Correct answer

B. (a−d)(c−a)(b−c)(d−b)\frac{(a-d)(c-a)}{(b-c)(d-b)}

Step-by-step solution

From, p⋅tan⁡x=q⋅tan⁡yp \cdot \tan x=q \cdot \tan yp2⋅tan⁡2x=q2⋅tan⁡2y⇒p2q2=tan⁡2ytan⁡2x=(d−a)(b−d)⋅(a−c)(c−b)=(a−d)(c−a)(b−c)(d−b)\begin{aligned} & p^2 \cdot \tan ^2 x=q^2 \cdot \tan ^2 y \\ & \Rightarrow \frac{p^2}{q^2}=\frac{\tan ^2 y}{\tan ^2 x}=\frac{(d-a)}{(b-d)} \cdot \frac{(a-c)}{(c-b)} \\ & \quad=\frac{(a-d)(c-a)}{(b-c)(d-b)} \end{aligned}$$
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