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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

If a ^ 2 x + b ^ 2 x = c, b ^ 2 y + a ^ 2 y = d and atan x = btan...

If 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 atanx=btany,atan x = btan y, then a2b2\dfrac{a^{2}}{b^{2}} is equal to (Here a,b,ca, b, c and& ddare distinct)

Options

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

Correct answer

B. (a−d)(c−a)(b−c)(d−b)\dfrac{\left(a - d\right)\left(c - a\right)}{\left(b - c\right)\left(d - b\right)}

Step-by-step solution

Dividing& the given equations by cos⁡2x and cos⁡2y\cos^{2} x \text{ and } \cos^{2} y respectively. We get,
tan⁡2x=c−ba−c\tan^{2}x = \dfrac{c - b}{a - c} and tan⁡2y=d−ab−d\tan^{2}y = \dfrac{d - a}{b - d}

⇒a2b2=tan⁡2ytan⁡2x=(a−d)(c−a)(b−c)(d−b)⇒ \dfrac{a2}{b2} = \dfrac{\tan^{2}y}{\tan^{2}x} = \dfrac{( a - d ) ( c - a )}{( b - c ) ( d - b )}

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