Question

If ${\theta }_1$ and ${\theta }_2$ are two solution of the equation $\alpha Cos2\theta +bsin\theta =c$ then $tan{\theta }_1+tan{\theta }_2=$

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

Answer 1

The correct option is A $\frac{2b}{c+a}$

$a\cos 2\theta +b\sin 2\theta =c$
$formulaused$
$\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$
$\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$
$a\cos 2\theta +b\sin 2\theta =c$
$\Rightarrow a\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)+b\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)=c$
$\Rightarrow a(1-{\tan }^2\theta )+2b\tan \theta =c(1+{\tan }^2\theta )$
$\Rightarrow {\tan }^2\theta (c+a)-2b\tan \theta +c-a=0$
$let{\theta }_1,{\theta }_{2}aretherootsoftheequation$
$sumofroots=\tan {\theta }_1+\tan {\theta }_2=\frac{-(-2b)}{c+a}=\frac{2b}{c+a}$