Question

If $tanx=\frac{2b}{a-c}(a\ne c),y=aco{s}^{2}x+2bsinxcosx+csi{n}^{2}xandz=asi{n}^{2}x-2bsinxcosx+co{s}^{2}x,$ then

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

Answer 1

The correct option is B $y+z=a+c$

We have,
$y+z=a(co{s}^{2}x+si{n}^{2}x)+c(si{n}^{2}x+co{s}^{2}x)=a+c$
$(\therefore solutionis(b\left)\right\}$
$y-z=a(co{s}^{2}x-si{n}^{2}x)+4bsinxcosx$
$-c(co{s}^{2}x-si{n}^{2}x)$
$=(a-c)cos2x+2bsin2x$
$=(a-c).\left(\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}\right)+2b.\left(\frac{2tanx}{1+ta{n}^{2}x}\right)$
$=(a-c).\left\{\frac{1-\frac{4b^2}{(a-c{)}^2}}{1+\frac{4b^2}{(a-c{)}^2}}\right\}+2b.\left\{\frac{2.\frac{2b}{(a-c)}}{1+\frac{4b^2}{(a-c{)}^2}}\right\}$
Since $tanx=\frac{2b}{(a-c)},$
$\therefore y-z=\frac{(a-c).\left\{\right(a-c{)}^2-4b^2\}+8b^2(a-c)}{(a-c{)}^2+4b^2}$
$=\frac{(a-c)(a-c{)}^2+4b^2}{\left\{\right(a-c{)}^2+4b^2\}}=(a-c)$
$\Rightarrow y\ne z,(\because a\ne c).$