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
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
- A
$y = z$
- B
$y + z = a + c$
- C
$y - z = a + c$
- D
$y - z = {(a - c)^2} + 4{b^2}$
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