2026/081) Given that $\frac{\sin (A-B)} {\sin(A+B)} =\frac{5}{13}$ How do you show that $4 \tan A = 9 \tan B$

We shall use the formula for sum of  sin and and difference of sin

$\sin \, C + \sin \, D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}\cdots(1)$

and  $\sin \, C - \sin \, D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}\cdots(2)$

We are given

$\frac{\sin (A-B)} {\sin(A+B)} =\frac{5}{13}$ 

Or   $\frac{\sin (A+B)} {\sin(A-B)} =\frac{13}{5}$

By compnendo and dividendo we get'

$\frac{\sin (A+B)+\sin(A-B)} {\sin(A+B) -\sin(A-B)} =\frac{13+5}{13-5}$

OR

$\frac{\sin (A+B)+\sin(A-B)} {\sin(A+B) -\sin(A-B)} =\frac{9}{4}\cdots(3)$  

we have  

$\sin (A+B)+\sin(A-B) = 2\sin\,A\cos\,B\cdots(4)$ using (1)

$\sin (A+B)-\sin(A-B) = 2\cos\,A\sin\,B\cdots(5)$ using (2)

Hence we have

$\frac{\sin (A+B)+\sin(A-B)} {\sin(A+B) -\sin(A-B)} =\frac{2\sin\,A\cos\,B}{2\cos\,A\sin\,B}$

Or  $\frac{\sin (A+B)+\sin(A-B)} {\sin(A+B) -\sin(A-B)} =\frac{\tan\,A}{\tan \,B}$ 

Form above and (3) 

$\frac{\tan\,A}{\tan\,B}=\frac{9}{4}$

Or  

$4 \tan A = 9 \tan B$