If a ^ 2 x + b ^ 2 x = c, b ^ 2 y + a ^ 2 y = d and atan x = btan y, then a^ 2 b^ 2 is equal to (Here a, b, c and d are distinct)

Correct answer: B. ((a - d)(c - a))/((b - c)(d - b)) Dividing the given equations by ^ 2 x & ^ 2 y respectively. We get, ^ 2 x = (c - b)/(a - c) and ^ 2 y = (d - a)/(b - d) ⇒ (a2)/(b2) = ^ 2 y ^ 2 x = (( a - d ) ( c - a ))/(( b - c ) ( d - b ))

JEE Main

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

JEE Main · Mathematics · Trigonometric Ratios & Identities

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

a\sin^{2} x + b\cos^{2} x = c, b\sin^{2} y + a\cos^{2} y = d

and

atan x = btan y,

then

\dfrac{a^{2}}{b^{2}}

is equal to (Here

a, b, c

and

d

are distinct)

Options

A $\dfrac{\left(b - c\right)\left(d - b\right)}{\left(a - d\right)\left(a - b\right)}$
B $\dfrac{\left(a - d\right)\left(c - a\right)}{\left(b - c\right)\left(d - b\right)}$
C $\dfrac{\left(d - a\right)\left(c - a\right)}{\left(b - c\right)\left(d - b\right)}$
D $\dfrac{\left(b - c\right)\left(b - d\right)}{\left(a - c\right)\left(a - d\right)}$

Correct answer

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 $\cos^{2} x & \cos^{2} y$ respectively. We get,
$\tan^{2}x = \dfrac{c - b}{a - c}$ and $\tan^{2}y = \dfrac{d - a}{b - d}$

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

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