(b) We have,

$y + z = a({\cos ^2}x + {\sin ^2}x) + c({\sin ^2}x + {\cos ^2}x) = a + c$

$(\therefore {\rm{solution\,\, is\,\, (b)}} )$

$y – z = a({\cos ^2}x – {\sin ^2}x) + 4b\sin x\cos x$

$ – c({\cos ^2}x – {\sin ^2}x)$

$ = (a – c)\cos 2x + 2b\sin 2x$

$ = (a – c).\,\left( {\frac{{1 – {{\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\}$

Since $\tan x = \frac{{2b}}{{(a – c)}}$,

$\therefore y – z = \frac{{(a – c).\{ {{(a – c)}^2} – 4{b^2}\} + 8{b^2}(a – c)}}{{{{(a – c)}^2} + 4{b^2}}}$

$ = \frac{{(a – c){{(a – c)}^2} + 4{b^2}}}{{\{ {{(a – c)}^2} + 4{b^2}\} }} = (a – c)$

$ \Rightarrow y \ne z\,,\,(a \ne c)$