Question

Let $\alpha ,\beta$ be two distinct roots of $a\cos \theta +b\sin \theta =c$, where $a,b$ and $c$ are three real constants and $\theta \in [0,2\pi ]$. Then $\alpha +\beta$ is also a root of the same equation if

• A. $a+b=c$
• B. $b+c=a$
• C. $c+a=b$
• D. $c=a$

Answer 1

The correct option is D $c=a$

We know,
$\cos \theta =\left(\frac{1-{\tan }^2\theta /2}{1+{\tan }^2\theta /2}\right)$
$\sin \theta =\left(\frac{2\tan \theta /2}{1+{\tan }^2\theta /2}\right)$
Substitute the value in the given equation.
$a\left(\frac{1-{\tan }^2\theta /2}{1+{\tan }^2\theta /2}\right)+b\left(\frac{2\tan \theta /2}{1+{\tan }^2\theta /2}\right)=c$
$\Rightarrow (c+a){\tan }^2\left(\theta /2\right)-2b\tan \left(\theta /2\right)+(c-a)=0$

Now, $\tan \alpha /2$ and $\tan \beta /2$ are roots of given equation.
$\therefore \tan \frac{\alpha }{2}+\tan \frac{\beta }{2}=\frac{2b}{c+a}$
and $\tan \frac{\alpha }{2}\cdot \tan \frac{\beta }{2}=\frac{c-a}{c+a}$

$\therefore \tan \frac{\alpha +\beta }{2}=\frac{\frac{2b}{c+a}}{1-\frac{c-a}{c+a}}=\frac{b}{a}$
$\frac{b}{a}$ is also a root of the equation,
$\Rightarrow (c+a){\left(\frac{b}{a}\right)}^2-2b\left(\frac{b}{a}\right)+c-a=0$
$\Rightarrow \frac{b}{a}=\frac{b\pm \sqrt{b^2-c^2+a^2}}{a+c}$

The equation hold true when $c=a$ and taking positive sign.