• # question_answer If $a\,\cos 2\theta +b\,\sin 2\theta =c$has a and b as its solution, then the value of $\tan \alpha +\tan \beta$ is [Kurukshetra CEE 1998] A) $\frac{c+a}{2b}$ B) $\frac{2b}{c+a}$ C) $\frac{c-a}{2b}$ D) $\frac{b}{c+a}$

$a\cos 2\theta +b\sin 2\theta =c$ Þ $a\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)+b\frac{2\tan \theta }{1+{{\tan }^{2}}\theta }=c$ $\Rightarrow$ $a-a{{\tan }^{2}}\theta +2b\tan \theta =c+c{{\tan }^{2}}\theta$ $\Rightarrow$$-(a+c){{\tan }^{2}}\theta +2b\,\tan \theta +(a-c)=0$ $\therefore \tan \alpha +\tan \beta =-\frac{2b}{-(c+a)}=\frac{2b}{c+a}$ .