Question

If $\tan \alpha ,\tan \beta ,\tan \gamma$ be the root of $a{t}^3+bt+c=0$, where $a,b$ and $cϵR$ and $cϵ\ne 0$. Also $\tan \alpha +\tan \beta =\lambda$ and $\alpha +\beta =\frac{\pi }{2}$ then the value of $\sin 2\alpha$ is

• A. $\frac{a}{c}$
• B. $\frac{2a}{c}$
• C. $\frac{c}{a}$
• D. $\frac{1}{2}\frac{c}{a}$

Answer 1

The correct option is B $\frac{2a}{c}$

$\tan \alpha +\tan \beta +\tan \gamma =0\Rightarrow \tan \gamma =-\lambda$
Also
$\tan \alpha \tan \beta \tan \gamma =-\frac{c}{a}$
$\Rightarrow \tan \alpha \tan \beta =\frac{c}{a\lambda }$
But
$\alpha +\beta =\frac{\pi }{2}\Rightarrow \frac{c}{a\lambda }=1\Rightarrow \lambda =\frac{c}{a}$

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

$\tan \alpha +\frac{1}{\tan \alpha }=\frac{c}{a}\Rightarrow \frac{2}{\sin 2\alpha }=\frac{c}{a}$

$\Rightarrow \sin 2\alpha =\frac{2a}{c}$