Question

If $\alpha$ and $\beta$ satisfies the equations $5\sin 2\beta =3\sin 2\alpha$ and $3\tan \alpha =\tan \beta$ simultaneously, where $0<\alpha <\pi ,0<\beta <\pi ,\alpha ,\beta \ne \frac{\pi }{2}$, then the possible value(s) of $\tan \alpha +\tan \beta$ is/are

• A. $4$
• B. $-4$
• C. $2$
• D. $-2$

Answer 1

Answer: (A), (B)

$-4$

$5\sin 2\beta =3\sin 2\alpha$
$3\tan \alpha =\tan \beta \cdots \left(1\right)$

$5\sin 2\beta =3\sin 2\alpha$
$\Rightarrow 5\frac{2\tan \beta }{1+{\tan }^2\beta }=3\frac{2\tan \alpha }{1+{\tan }^2\alpha }$
Using equation $\left(1\right)$,
$\Rightarrow 5\frac{3\tan \alpha }{1+9{\tan }^2\alpha }=3\frac{\tan \alpha }{1+{\tan }^2\alpha }\Rightarrow \frac{5}{1+9{\tan }^2\alpha }=\frac{1}{1+{\tan }^2\alpha }(\because \tan \alpha \ne 0)$
$\Rightarrow 5+5{\tan }^2\alpha =1+9{\tan }^2\alpha$
$\Rightarrow {\tan }^2\alpha =1$
$\Rightarrow \tan \alpha =\pm 1$

When $\tan \alpha =1$
$3\tan \alpha =\tan \beta \Rightarrow \tan \beta =3\Rightarrow \tan \alpha +\tan \beta =1+3=4$

When $\tan \alpha =-1$
$\tan \beta =-3\Rightarrow \tan \alpha +\tan \beta =-1-3=-4$