Question

If $\tan x=\frac{2b}{a-c}(a\ne c),y=a{\cos }^{2}x+2b\sin x\times \cos x+c{\sin }^{2}x$ 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. $y-z=(a-c{)}^2+4b^2$

Answer 1

Answer: (B), (C)

$y-z=a-c$
$y+z=a+c$

$y=a{\cos }^{2}x+2b\sin x\cos x+c{\sin }^{2}x$ ...(1)

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

Adding (1) and (2), we get

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

Subtracting (2) from (1) we get

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

Substituting $\tan x=\frac{2b}{(a-c)}$

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

Hence, options 'B' and 'C' are correct.