Question

If 3 sin beta=sin(2 alpha+beta), then tan (alpha +beta)-2tan alpha is

Answer 1

Please see the attachment

Answer 2

Answer:

$\tan(\alpha + \beta)-2\tan \alpha=0$

Step-by-step explanation:

We need to find $\tan (\alpha + \beta)-2\tan \alpha$

Given:- $3\sin \beta= \sin(2 \alpha + \beta)$

As, $3\sin \beta= \sin(2 \alpha + \beta)$

Divide both the sides by $\sin \beta$

$3= \frac{\sin(2 \alpha + \beta)}{\sin \beta}$

apply componendo and dividendo,

$\frac{3+1}{3-1}= \frac{\sin(2 \alpha + \beta)+\sin \beta}{\sin(2 \alpha + \beta)-\sin \beta}$

Since, $\sin \alpha +\sin \beta = 2\sin(\frac{\alpha + \beta}{2})\cos(\frac{\alpha - \beta}{2})$

and $\sin \alpha -\sin \beta = 2\cos(\frac{\alpha + \beta}{2})\sin(\frac{\alpha - \beta}{2})$

$2= \frac{2\sin(\frac{2 \alpha + \beta+ \beta}{2})\cos \frac{(2 \alpha + \beta - \beta)}{2}}{2\cos(\frac{2 \alpha + \beta+ \beta}{2})\sin \frac{(2 \alpha + \beta - \beta)}{2}}$

$2= \frac{\sin(\alpha + \beta)\cos \alpha}{\cos(\alpha + \beta)\sin \alpha}$

$2= \frac{\tan(\alpha + \beta)}{\tan \alpha}$

Multiply both the sides by $\tan \alpha$

$2\tan \alpha = \tan(\alpha + \beta)$

Subtract both the sides by $2\tan \alpha$

$\tan(\alpha + \beta)-2\tan \alpha=0$

Hence, $\tan(\alpha + \beta)-2\tan \alpha=0$