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

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(b) $a{\sin ^2}x + b{\cos ^2}x = c \Rightarrow (b - a){\cos ^2}x = c - a$ 

$\Rightarrow (b - a) = (c - a)(1 + {	an ^2}x)$

$b{\sin ^2}y + a

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) $a{\sin ^2}x + b{\cos ^2}x = c \Rightarrow (b - a){\cos ^2}x = c - a$ 

$\Rightarrow (b - a) = (c - a)(1 + {	an ^2}x)$

$b{\sin ^2}y + a{\cos ^2}y = d \Rightarrow (a - b){\cos ^2}y = d - b$ 

$ \Rightarrow (a - b) = (d - b)(1 + {	an ^2}y)$

$	herefore {	an ^2}x = \frac{{b - c}}{{c - a}},\,\,{	an ^2}y = \frac{{a - d}}{{d - b}}$

$	herefore \frac{{{{	an }^2}x}}{{{{	an }^2}y}} = \frac{{(b - c)(d - b)}}{{(c - a)(a - d)}}$&hellip;..$(i)$

But $a	an x = b	an y,$

$i.e.,$ $\frac{{	an x}}{{	an y}} = \frac{b}{a}$&hellip;..$(ii)$ 

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)}}$.

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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