If a sin^2 x + b cos^2 x = c,b sin^2 y + a co | vedclass

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(B) Given $a \sin^2 x + b \cos^2 x = c$. Dividing by $\cos^2 x$,we get $a 	an^2 x + b = c \sec^2 x = c(1 + 	an^2 x)$.So,$(a - c) 	an^2 x = c - b$,w

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$\frac{(a - d)(c - a)}{(b - c)(d - b)}$

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(B) Given $a \sin^2 x + b \cos^2 x = c$. Dividing by $\cos^2 x$,we get $a 	an^2 x + b = c \sec^2 x = c(1 + 	an^2 x)$.So,$(a - c) 	an^2 x = c - b$,which implies $	an^2 x = \frac{b - c}{a - c} = \frac{c - b}{c - a}$.Similarly,for $b \sin^2 y + a \cos^2 y = d$,dividing by $\cos^2 y$ gives $b 	an^2 y + a = d(1 + 	an^2 y)$.So,$(b - d) 	an^2 y = d - a$,which implies $	an^2 y = \frac{d - a}{b - d} = \frac{a - d}{d - b}$.Given $a 	an x = b 	an y$,we have $\frac{	an x}{	an y} = \frac{b}{a}$.Squaring both sides,$\frac{	an^2 x}{	an^2 y} = \frac{b^2}{a^2}$.Substituting the values,$\frac{b^2}{a^2} = \frac{(c - b)/(c - a)}{(a - d)/(d - b)} = \frac{(b - c)(d - b)}{(c - a)(a - d)}$.Therefore,$\frac{a^2}{b^2} = \frac{(c - a)(a - d)}{(b - c)(d - b)}$.

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

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