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$a \sin^{2} x + b \cos^{2} x = c, b \sin^{2} y + a \cos^{2} y = d and a \tan x = b \tan y then \frac{a^{2}}{b^{2}} =$

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b−cd−ba−dd−b

a−dc−ab−cd−b

d−ac−ab−cd−b

b−ab−da−ca−d

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detailed solution

Correct option is B

We haveasin2x+bcos2x=c⇒atan2x+b=csec2x⇒tan2x=c-ba-csimilarly tan2y=d-ab-d∴a2b2=tan2ytan2x=(a-d)(c-a)(b-c)(d-b)

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asin2x+bcos2x=c, bsin2y+acos2y=d and atanx=btany then a2b2=

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$a \sin^{2} x + b \cos^{2} x = c, b \sin^{2} y + a \cos^{2} y = d and a \tan x = b \tan y then \frac{a^{2}}{b^{2}} =$