Question

If $x_1$ and $x_2$ are two distinct roots of the equation $a\cos x+b\sin x=c,$ then $\tan \frac{x_1+x_2}{2}$ is equal to :

• A. $\frac{a}{b}$
• B. $\frac{b}{a}$
• C. $\frac{c}{a}$
• D. $\frac{a}{c}$

Answer 1

The correct option is B $\frac{b}{a}$

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

$\Rightarrow (c+a){\tan }^2\left(\frac{x}{2}\right)-2b\tan \left(\frac{x}{2}\right)+(c-a)=0$

$\Rightarrow \tan \left(\frac{x_1}{2}\right)+\tan \left(\frac{x_2}{2}\right)=\frac{2b}{c+a}$
and
$\tan \left(\frac{x_1}{2}\right).\tan \left(\frac{x_2}{2}\right)=\frac{c-a}{c+a}$
Hence,
$\tan \left(\frac{x_1+x_2}{2}\right)=\frac{\tan \left(\frac{x_1}{2}\right)+\tan \left(\frac{x_2}{2}\right)}{1-\tan \left(\frac{x_1}{2}\right).\tan \left(\frac{x_2}{2}\right)}=\frac{\frac{2b}{c+a}}{1-\frac{c-a}{c+a}}=\frac{b}{a}$