Question

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

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

Answer 1

Answer: (C)

$y-z=a-c$

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

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

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

$y+z=a+c$

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

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

$=(a-c)[1-2{\sin }^{2}x+\left(\frac{2b}{a-c}\right)2\sin x\cos x]$

$=(a-c)[1-2{\sin }^{2}x+\tan x2\sin x\cos x]$

$=(a-c)(1-2{\sin }^{2}x+2{\sin }^{2}x)$

$(y-z)=(a-c)$