Question

If $a\cos 2\theta +b\sin 2\theta =c$ as $\alpha$ and $\beta$ as solution then the value of $\tan \alpha +\tan \beta is$

• A. $\frac{c+a}{2b}$
• B. $\frac{2b}{c+a}$
• C. $\frac{c-a}{2b}$
• D. $\frac{b}{c+a}$

Answer 1

Answer: (B)

$\frac{2b}{c+a}$

$a\cos 2\theta +b\sin 2\theta =c$

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

$\Rightarrow a-a{\tan }^2\theta +2b\tan \theta -c(1+{\tan }^2\theta )=0$

$\Rightarrow -a{\tan }^2\theta +2b\tan \theta -c(1+{\tan }^2\theta )+a=0$

$\Rightarrow -a{\tan }^2\theta -2b\tan \theta +c(1+{\tan }^2\theta )-a=0$

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

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

Let $\tan \alpha$ and $\tan \beta$ be the roots of the equation then

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