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)

$c=a$

Given equation is $a\cos \theta +b\sin \theta =c$

$\Rightarrow a\left(\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\right)+\frac{2b{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=c$

$[\because \cos \theta =\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}and\sin \theta =\frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}]$

$\Rightarrow a(1-{\tan }^2\frac{\theta }{2})+2b\tan \frac{\theta }{2}=c(1+{\tan }^2\frac{\theta }{2})$

$\Rightarrow a-a{\tan }^2\frac{\theta }{2}+2b\tan \frac{\theta }{2}-c-c{\tan }^2\frac{\theta }{2}=0$

$\Rightarrow (c+a){\tan }^2\frac{\theta }{2}-2b\tan \frac{\theta }{2}+(c-a)=0$

Let $\alpha$ and $\beta$ be the roots of the equation.

$\alpha +\beta =\frac{2b}{c+a}$ 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.

$\therefore (c+a)\frac{b^2}{a^2}-2b\left(\frac{b}{a}\right)+c-a=0$

$\Rightarrow (c+a)\frac{b^2}{a^2}-2b\left(\frac{b}{a}\right)+c-a=0$

$\Rightarrow b^{2}c+b^{2}a-2b^{2}a+c{a}^2-a^3=0$

$\Rightarrow b^{2}a+b^{2}c+c{a}^2-a^3=0$

$\Rightarrow b^{2}c-b^{2}a+c{a}^2-a^3=0$

$\Rightarrow b^2(c-a)+a^2(c-a)=0$

$(c-a)(b^2+a^2)=0$

$\Rightarrow c-a=0$ or $b^2+a^2=0$

$\Rightarrow c=a$ or $b^2+a^2=0$