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