Question

Let $\alpha$ and $\beta$ be two distinct roots of $a\cos \theta +b\sin \theta =c$, where $a,b,c$ 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

Answer: (D)

Given equation is $a\cos \theta +b\sin \theta =c$
$\Rightarrow \alpha \left(\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\right)+\frac{2b\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=c$
$[\because \cos \theta =\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}$ and $\sin \theta =\frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}]$
$\Rightarrow a\left(1-{\tan }^2\frac{\theta }{2}\right)+2b\tan \frac{\theta }{2}$
$=c\left(1+{\tan }^2\frac{\theta }{2}\right)$
$\Rightarrow a-ata{n}^2\frac{\theta }{2}+2b\tan \frac{\theta }{2}-c-c{\tan }^2\frac{\theta }{2}=0$
$\Rightarrow (c+a){\tan }^2\frac{\theta }{2}-2b\tan \frac{\theta }{2}+(c-a)=0$
Let $\alpha$ and $\beta$ be the roots of the equation.
$\therefore \alpha +\beta =\frac{2b}{c+a}$ and $\alpha \beta =\frac{c-a}{c+a}$
Now, $\tan \frac{\alpha +\beta }{2}=\frac{\frac{2b}{c+a}}{1-\frac{c-a}{c+a}}$
$=\frac{\frac{2b}{c+a}}{\frac{c+a-c+a}{c+a}}=\frac{b}{a}$
Since, $\frac{b}{a}$ is a root of the equation.
$\therefore (c+a)\frac{b^2}{a^2}-2b\left(\frac{b}{a}\right)+c-a=0$
$\Rightarrow b^{2}c+b^{2}a-2b^{2}a+c{a}^2-a^3=0$
$\Rightarrow -b^{2}a+b^{2}c+c{a}^2-a^3=0$
$\Rightarrow b^{2}c-b^{2}a+c{a}^2-a^3=0$
$\Rightarrow b^2(c-a)+a^2(c-a)=0$
$\Rightarrow (c-a)\left(b^2+a^2\right)=0$
$\Rightarrow c-a=0$ or $b^2+a^2=0$
$\Rightarrow c=a$ or $b^2+a^2=0$