If x₁ and x₂ are two distinct roots of the equation $acos x + bsin x = c,$ then $\tan \frac{x_{1} + x_{2}}{2}$ is equal to

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$\frac{a}{b}$

$\frac{b}{a}$

$\frac{c}{a}$

$\frac{a}{c}$

answer is B.

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Detailed Solution

$\begin{array}{l} acos x + bsin x = c \\ \Rightarrow \frac{a (1 - \tan^{2} \frac{x}{2})}{1 + \tan^{2} \frac{x}{2}} + \frac{2 btan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} = c \\ \Rightarrow (c + a) \tan^{2} \frac{x}{2} - 2 btan \frac{x}{2} + c - a = 0 \\ \Rightarrow \tan \frac{x_{1}}{2} + \tan \frac{x_{2}}{2} = \frac{2 b}{c + a} \\ and \tan \frac{x_{1}}{2} \tan \frac{x_{2}}{2} = \frac{c - a}{c + a} \\ \Rightarrow \tan (\frac{x_{1} + x_{2}}{2}) = \frac{\frac{2 b}{c + a}}{1 - \frac{c - a}{c + a}} = \frac{2 b}{2 a} = \frac{b}{a} \end{array}$