Question

lf $\tan x=\frac{2b}{a-c}$ , $a\ne c$ and $y=a{\cos }^{2}x+2b\sin x\cos x+c{\sin }^{2}x$, $z=a{\sin }^{2}x-2b\sin x\cos x+c{\cos }^{2}x$, then

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

Answer 1

Answer: (C)

$y-z=a-c$

$y=a{\cos }^{2}x+2b\sin x\cos x+c{\sin }^{2}xz=a{\sin }^{2}x-2b\sin x\cos x+c{\cos }^{2}x..\left(2\right)y-z=(a-c)\cos 2x+2b\sin 2x$ ...(1)
As $\tan x=\frac{2b}{a-c}$, we get
$\cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}=\frac{{(a-c)}^2-4b^2}{{(a-c)}^2+4b^2}\sin 2x=\frac{2\tan x}{1+{\tan }^{2}x}=\frac{4b(a-c)}{{(a-c)}^2+4b^2}$
Substituting this in (1) and (2) we get
$y-z=a-c$