Question

If $\alpha ,\beta$ are different values of x satisfying a cos x + b sin x = c then $tan\left(\frac{\alpha +\beta }{2}\right)=$

• A. a + b
• B. a – b
• C.
• D.

Answer 1

The correct option is D

$acosx+bsinx=c\Rightarrow a\left[\frac{1-ta{n}^{2}x/2}{1+ta{n}^{2}x/2}\right]+b\left[\frac{2tanx/2}{1+ta{n}^{2}x/2}\right]=c$
$\Rightarrow a\left(\frac{1-t^2}{1+t^2}\right)+\left(\frac{2t}{1+t^2}\right)=c,wheret=tan\frac{x}{2}$
$\Rightarrow a(1-t^2)+2bt=c(1+t^2)\Rightarrow (c+a)t^2-2bt+(c-a)=\mathrm{0....}\left(1\right)$
Since $\alpha ,\beta$ are the values of x, we get $tan\frac{\alpha }{2},tan\frac{\beta }{2}$ are the roots of (1).
$\therefore tan\frac{\alpha }{2}+tan\frac{\beta }{2}=\frac{2b}{c+a},tan\frac{\alpha }{2},tan\frac{\beta }{2}=\frac{c-a}{c+a}$
$tan\left(\frac{\alpha +\beta }{2}\right)=\frac{tan\alpha /2+tan\beta /2}{1-tan\left(\alpha /2\right)tan\left(\beta /2\right)}=\frac{\left(2b\right)/(c+a)}{1-(c-a)/(c+a)}=\frac{2b}{c+a-c+a}=\frac{2b}{2a}=\frac{b}{a}$