If a sin^2 x+b cos^2 x = c, b sin^2 y+a cos^2 y = d and a tan^2 x = b tan y then (a^2 /b^2 ) is equal to


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Question Number 18702 by Christo01 last updated on 28/Jul/17

$If a \sin^{2} x + b \cos^{2} x = c,$ $b \sin^{2} y + a \cos^{2} y = d and$ $a \tan^{2} x = b \tan y$ $then \frac{a^{2}}{b^{2}} is equal to$
Commented by mondodotto@gmail.com last updated on 29/Jul/17

$please help me to solve this question$
	Answered by behi.8.3.4.1.7@gmail.com last updated on 30/Jul/17
	$a.tg^2 x+b=c+c.tg^2 x⇒tg^2 x=((c−b)/(a−c)) b.tg^2 y+a=d+d.tg^2 y⇒tg^2 y=((d−a)/(b−d)) a.tg^2 x=b.tg^2 y⇒a.((c−b)/(a−c))=b.((d−a)/(b−d))⇒ ⇒a(bc−dc−b^2 +bd)=b(ad−a^2 −dc+ac) abc−adc−ab^2 +abd=abd−a^2 b−bcd+abc a^2 b−ab^2 =adc−bdc ab(a−b)=cd(a−b)⇒ { ((a=b⇒(a^2 /b^2 )=1)),((ab=cd)) :}$
	$a . {t g}^{2} x + b = c + c . {t g}^{2} x \Rightarrow {t g}^{2} x = \frac{c - b}{a - c}$ $b . {t g}^{2} y + a = d + d . {t g}^{2} y \Rightarrow {t g}^{2} y = \frac{d - a}{b - d}$ $a . {t g}^{2} x = b . {t g}^{2} y \Rightarrow a . \frac{c - b}{a - c} = b . \frac{d - a}{b - d} \Rightarrow$ $\Rightarrow a (b c - d c - b^{2} + b d) = b (a d - a^{2} - d c + a c)$ $a b c - a d c - {a b}^{2} + a b d = a b d - a^{2} b - b c d + a b c$ $a^{2} b - {a b}^{2} = a d c - b d c$ $a b (a - b) = c d (a - b) \Rightarrow {\begin{cases} a = b \Rightarrow \frac{a^{2}}{b^{2}} = 1 \\ a b = c d \end{cases}$
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