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# If $a\cos 2\theta+b\sin 2\theta=c$ has $\alpha$ & $\beta$ as its solutions then $\tan\alpha+\tan\beta$ and $\tan\alpha\tan\beta$ is equal to

$\begin{array}{1 1}(a)\;\large\frac{2b}{c+a},\frac{c-a}{c+a}&(b)\;\large\frac{c+a}{2b},\frac{c-a}{c+a}\\(c)\;\large\frac{2b}{c-a},\frac{c-a}{c+a}&(d)\;\large\frac{2b}{c+a},\frac{c+a}{c-a}\end{array}$

We have $a\cos 2\theta+b\sin 2\theta=c$
$\Rightarrow a(\cos^2\theta-\sin^2\theta)+2b\sin\theta\cos c=c$
$\Rightarrow a(1-\tan^2\theta)+2b\tan\theta=c\sec^2\theta$
$\qquad\qquad\qquad\qquad\qquad\;=c(1+\tan^2\theta)$
$\Rightarrow \tan^2\theta(c+a)-2b\tan\theta+c+a=0$
This equation has $\tan\alpha$ and $\tan\beta$ as its roots.
$\Rightarrow \tan\alpha+\tan\beta=\large\frac{2b}{c+a}$
$\tan\alpha\tan\beta=\large\frac{c-a}{c+a}$
Hence (a) is the right option.