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यदि $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$
$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$
तथा $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ हो, तब
$y = z$
$y + z = a + c$
$y - z = a + c$
$y - z = {(a - c)^2} + 4{b^2}$
Solution
(b) यहाँ , $y + z = a({\cos 4^2}x + {\sin ^2}x) + c({\sin ^2}x + {\cos ^2}x) = a + c$
$(\therefore (b) $ हल है $)$
$y – z = a({\cos ^2}x – {\sin ^2}x) + 4b\sin x\cos x$
$ – c({\cos ^2}x – {\sin ^2}x)$
$ = (a – c)\cos 2x + 2b\sin 2x$
$ = (a – c).\,\left( {\frac{{1 – {{\tan }^2}x}}{{1 + {{\tan }^2}x}}} \right) + 2b.\left( {\frac{{2\tan x}}{{1 + {{\tan }^2}x}}} \right)$
$ = (a – c).\left\{ {\frac{{1 – 4{b^2}/{{(a – c)}^2}}}{{1 + 4{b^2}/{{(a – c)}^2}}}} \right\} + 2b.\left\{ {\frac{{2.2b/(a – c)}}{{1 + 4{b^2}/{{(a – c)}^2}}}} \right\}$
चूँकि $\tan x = \frac{{2b}}{{(a – c)}}$,
$\therefore y – z = \frac{{(a – c).\{ {{(a – c)}^2} – 4{b^2}\} + 8{b^2}(a – c)}}{{{{(a – c)}^2} + 4{b^2}}}$
$ = \frac{{(a – c){{(a – c)}^2} + 4{b^2}}}{{\{ {{(a – c)}^2} + 4{b^2}\} }} = (a – c)$
$ \Rightarrow y \ne z\,,\,(a \ne c)$
