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

$\frac{b}{a}$

Explanation for correct option:

Evaluate the value of the given trigonometric expression

Given, two distinct roots of $\mathrm{a}\cos \mathrm{x}+\mathrm{b}\sin \mathrm{x}=\mathrm{c}$ are ${\mathrm{x}}_1\mathrm{and}{\mathrm{x}}_2$

We know that, $\cos \mathrm{x}=\frac{1-{\tan }^2\left({\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}{1+{\tan }^2\left({\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}$ and $\sin 2\mathrm{x}=\frac{2\tan {\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{1+{\tan }^2\left({\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}$

Replacing these values in the given equation, we get,

$$\begin{gathered} \Rightarrow \mathrm{a}\left[\frac{(1-{\tan }^2({\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left)\right)}{(1+{\tan }^2({\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left)\right)}\right]+\mathrm{b}[\frac{2\tan {\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{(1+{\tan }^2({\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left)\right)}]=\mathrm{c} \\ \Rightarrow \mathrm{a}–\mathrm{a}{\tan }^2\left(\frac{\mathrm{x}}{2}\right)+2\mathrm{b}\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{c}+\mathrm{c}{\tan }^2\left(\frac{\mathrm{x}}{2}\right) \\ \Rightarrow \mathrm{c}+\mathrm{c}{\tan }^2\left(\frac{\mathrm{x}}{2}\right)-\mathrm{a}+\mathrm{a}{\tan }^2\left(\frac{\mathrm{x}}{2}\right)-2\mathrm{b}\tan \left(\frac{\mathrm{x}}{2}\right)=0 \\ \Rightarrow (\mathrm{a}+\mathrm{c}){\tan }^2\left(\frac{\mathrm{x}}{2}\right)–2\mathrm{b}\tan \left(\frac{\mathrm{x}}{2}\right)+\mathrm{c}-\mathrm{a}=0 \end{gathered}$$

Since, $x_1$ and $x_2$ are the roots of the equation, so, for the above quadratic equation we can say that,

Sum of roots $=\tan \left(\raisebox{1ex}{$x_1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)+\tan \left(\raisebox{1ex}{$x_2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=\frac{2b}{c-a}\mathbf{}\left[\because \mathrm{sum}\mathrm{of}\mathrm{roots}=-\frac{b}{a}\right]$

and, Product of roots $\tan \left(\raisebox{1ex}{$x_1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\times \tan \left(\raisebox{1ex}{$x_2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=\frac{c-a}{c+a}\mathbf{}\left[\because \mathrm{Product}\mathrm{of}\mathrm{roots}=\frac{c}{a}\right]$

Thus,

$$\begin{gathered} \tan \frac{(x_1+x_2)}{2}=\frac{\tan \left(\raisebox{1ex}{$x_1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)+\tan \left(\raisebox{1ex}{$x_2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}{1-\tan \left(\raisebox{1ex}{$x_1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\times \tan \left(\raisebox{1ex}{$x_2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)} \\ =\frac{{\displaystyle \frac{2b}{c+a}}}{1-{\displaystyle \frac{c-a}{c+a}}} \\ =\frac{2b}{c+a-c+a} \\ =\frac{2b}{2a} \\ =\frac{b}{a} \end{gathered}$$

Hence, the answer is $\frac{b}{a}$, option (B) is correct.