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If $a \cos 2 \theta+b \sin 2 \theta=c$ has $\alpha$ and $\beta$ as its roots, then prove that $\tan \alpha+\tan \beta=\frac{2 b}{a+c}$.
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Verified Answer
Given that, $a \cos 2 \theta+b \sin 2 \theta=c$
$$
\begin{aligned}
&\Rightarrow a\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)+b\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=c \\
&{\left[\because \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta} \text { and } \cos 2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right]} \\
&\Rightarrow a\left(1-\tan ^2 \theta+2 b \tan \theta=c\left(1+\tan ^2 \theta\right)\right. \\
&\Rightarrow a-a \tan ^2 \theta+2 b \tan \theta=c+c \tan ^2 \theta \\
&\Rightarrow(a+c) \tan ^2 \theta-2 b \tan \theta+c-a=0
\end{aligned}
$$
Since, this equation has $\tan \alpha$ and $\tan \beta$ as its roots.
$$
\therefore \quad \tan \alpha+\tan \beta=\frac{-(-2 b)}{a+c}=\frac{2 b}{a+c}
$$
Hence proved.
$$
\begin{aligned}
&\Rightarrow a\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)+b\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=c \\
&{\left[\because \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta} \text { and } \cos 2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right]} \\
&\Rightarrow a\left(1-\tan ^2 \theta+2 b \tan \theta=c\left(1+\tan ^2 \theta\right)\right. \\
&\Rightarrow a-a \tan ^2 \theta+2 b \tan \theta=c+c \tan ^2 \theta \\
&\Rightarrow(a+c) \tan ^2 \theta-2 b \tan \theta+c-a=0
\end{aligned}
$$
Since, this equation has $\tan \alpha$ and $\tan \beta$ as its roots.
$$
\therefore \quad \tan \alpha+\tan \beta=\frac{-(-2 b)}{a+c}=\frac{2 b}{a+c}
$$
Hence proved.
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