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

Answer: (B)

$\frac{b}{a}$

Given, $a\cos x+b\sin x=c$
$\Rightarrow a\frac{(1-{\tan }^2\frac{x}{2})}{(1+{\tan }^2\frac{x}{2})}+\frac{2b\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=c$
$\Rightarrow (c+a){\tan }^2\frac{x}{2}-2b\tan \frac{x}{2}+c-a=0$
$\Rightarrow \tan \frac{x_1}{2}+\tan \frac{x_2}{2}=\frac{2b}{c+a}$
$\tan \frac{x_1}{2}\cdot \tan \frac{x_2}{2}=\frac{c-a}{c+a}$
$\Rightarrow \tan \left(\frac{x_1+x_2}{2}\right)=\frac{\frac{2b}{c+a}}{1-\frac{c-a}{c+a}}$
$=\frac{2b}{2a}=\frac{b}{a}$