If tan x = frac{{2b}}{{a - c}}(a ne c),

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Question

(b) We have,

$y + z = a({\cos ^2}x + {\sin ^2}x) + c({\sin ^2}x + {\cos ^2}x) = a + c$

$(	herefore {m{solution\,\, is\,\, (b)}} )$

$y - z

Vedclass · Accepted Answer

$y + z = a + c$

Vedclass · Answer

(b) We have,

$y + z = a({\cos ^2}x + {\sin ^2}x) + c({\sin ^2}x + {\cos ^2}x) = a + c$

$(	herefore {m{solution\,\, is\,\, (b)}} )$

$y - z = a({\cos ^2}x - {\sin ^2}x) + 4b\sin x\cos x$

$ - c({\cos ^2}x - {\sin ^2}x)$

$ = (a - c)\cos 2x + 2b\sin 2x$

$ = (a - c).\,\left( {\frac{{1 - {{	an }^2}x}}{{1 + {{	an }^2}x}}} ight) + 2b.\left( {\frac{{2	an x}}{{1 + {{	an }^2}x}}} ight)$

$ = (a - c).\left\{ {\frac{{1 - 4{b^2}/{{(a - c)}^2}}}{{1 + 4{b^2}/{{(a - c)}^2}}}} ight\} + 2b.\left\{ {\frac{{2.2b/(a - c)}}{{1 + 4{b^2}/{{(a - c)}^2}}}} ight\}$

Since $	an x = \frac{{2b}}{{(a - c)}}$,

$	herefore 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 
e z\,,\,(a 
e c)$

If $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$

$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$

and $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ then

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