(b) $a{\sin ^2}x + b{\cos ^2}x = c \Rightarrow (b – a){\cos ^2}x = c – a$ 

$\Rightarrow (b – a) = (c – a)(1 + {\tan ^2}x)$

$b{\sin ^2}y + a{\cos ^2}y = d \Rightarrow (a – b){\cos ^2}y = d – b$ 

$ \Rightarrow (a – b) = (d – b)(1 + {\tan ^2}y)$

$\therefore {\tan ^2}x = \frac{{b – c}}{{c – a}},\,\,{\tan ^2}y = \frac{{a – d}}{{d – b}}$

$\therefore \frac{{{{\tan }^2}x}}{{{{\tan }^2}y}} = \frac{{(b – c)(d – b)}}{{(c – a)(a – d)}}$…..$(i)$

But $a\tan x = b\tan y,$

$i.e.,$ $\frac{{\tan x}}{{\tan y}} = \frac{b}{a}$…..$(ii)$ 

From $(i)$ and $(ii), $

$\frac{{{b^2}}}{{{a^2}}} = \frac{{(b – c)(d – b)}}{{(c – a)(a – d)}}$

$ \Rightarrow \frac{{{a^2}}}{{{b^2}}} = \frac{{(c – a)(a – d)}}{{(b – c)(d – b)}}$.