• question_answer 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) $y-z={{(a-c)}^{2}}+4{{b}^{2}}$

We have, $y+z=a({{\cos }^{2}}x+{{\sin }^{2}}x)+c({{\sin }^{2}}x+{{\cos }^{2}}x)=a+c$ $(\therefore \text{solution is (b)}\}$ $y-z=a({{\cos }^{2}}x-{{\sin }^{2}}x)+4b\sin x\cos x$$-c({{\cos }^{2}}x-{{\sin }^{2}}x)$ $=(a-c)\cos 2x+2b\sin 2x$ $=(a-c).\,\left( \frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x} \right)+2b.\left( \frac{2\tan x}{1+{{\tan }^{2}}x} \right)$ $=(a-c).\left\{ \frac{1-4{{b}^{2}}/{{(a-c)}^{2}}}{1+4{{b}^{2}}/{{(a-c)}^{2}}} \right\}+2b.\left\{ \frac{2.2b/(a-c)}{1+4{{b}^{2}}/{{(a-c)}^{2}}} \right\}$ Since $\tan x=\frac{2b}{(a-c)}$, $\therefore y-z=\frac{(a-c).\{{{(a-c)}^{2}}-4{{b}^{2}}\}+8{{b}^{2}}(a-c)}{{{(a-c)}^{2}}+4{{b}^{2}}}$ $=\frac{(a-c){{(a-c)}^{2}}+4{{b}^{2}}}{\{{{(a-c)}^{2}}+4{{b}^{2}}\}}=(a-c)$ $\Rightarrow y\ne z\,,\,(\because a\ne c)$.