Question

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

The correct options are
B $y+z=a+c$
C $y-z=a-c$

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

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

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

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

$=a+c$

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

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

$=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}(a-c)+2b\frac{2\tan x}{1+{\tan }^{2}x}$

$=\frac{1-{\left(\frac{2b}{a-c}\right)}^2}{1+{\left(\frac{2b}{a-c}\right)}^2}(a-c)+2b\frac{2\frac{2b}{a-c}}{1+{\left(\frac{2b}{a-c}\right)}^2}=a-c$