Find $\frac{\tan(x+y)}{\tan(x)}$ if $5\sin(2x+y)=7\sin(y)$

Question

I saw this problem Find $\frac{\tan(x+y)}{\tan(x)}$ if $5\sin(2x+y)=7\sin(y)$

I tried to use $$\sin(x+y)\cos(x)+\cos(x+y)\sin(x)= \frac{7}{5}\sin(y)$$
$$\tan(x+y)=\frac{\frac{7}{5}\frac{\sin(y)}{\cos(x+y)}- \sin(x) }{\cos(x)}$$

$$\frac{\tan(x+y)}{\tan(x)}={\frac{7}{5}\frac{\sin(y)}{\sin(x)\cos(x+y)}- 1 }$$

Which lead to nothing.

I also tried to use $\tan(a+b)$ formula but it also didn't lead to anything useful.

Answer 1

Let $w = x + y$ , then the given equation becomes:

$5 \sin( x + w ) = 7 \sin(w - x) $

and we want $\tan(w)/\tan x$. Expanding:

$5 ( \sin x \cos w + \cos x \sin w ) = 7 (\sin w \cos x - \cos w \sin x )$

So,

$12 \sin x \cos w = 2 \sin w \cos x$

which gives:

$12 \tan x = 2 \tan w $

Therefore,

$\dfrac{\tan w}{\tan x} = 6$