Question

If $\alpha$ and $\beta$ are the roots of the equation $a\cos 2\theta +b\sin 2\theta =c$, then $\tan (\alpha +\beta )$ is equal to:

• A. $\frac{a}{b}$
• B. $\frac{-b}{a}$
• C. $\frac{b}{a}$
• D. $\frac{-a}{b}$

Answer 1

The correct option is C $\frac{b}{a}$

We have,

$a\cos 2\theta +b\sin 2\theta =c.............\left(i\right)$

and we know that, $\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$

$a\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }+b\frac{2\tan \theta }{1+{\tan }^2\theta }=c\Rightarrow {\tan }^2\theta (c+a)-2b\tan \theta +c-a=\mathrm{0.............}\left(ii\right)$

\sin ce, Equation (i) has roots $\alpha ,\beta$

so, Equation (ii) has roots $\tan \alpha ,\tan \beta$

Hence, $\tan \alpha +\tan \beta =\frac{2b}{c+a}\tan \alpha \tan \beta =\frac{c-a}{c+a}$

$\tan (\alpha +\beta )=\frac{\frac{2b}{c+a}}{1-(\frac{c-a}{c+a})}=\frac{2b}{2a}=\frac{b}{a}$

Therefore, Answer is $\frac{b}{a}$