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If $\large\frac{\sin(x+y)}{\sin(x-y)}=\frac{a+b}{a-b}$ then show that $\large\frac{\tan x}{\tan y}=\frac{a}{b}$

1 Answer

Toolbox:
  • $\sin A+\sin B=2\sin(\large\frac{A+B}{2})$$.\cos (\large\frac{A-B}{2})$
  • $\sin A-\sin B=2\cos(\large\frac{A+B}{2})$$.\sin(\large\frac{A-B}{2})$
Given :
$\large\frac{\sin(x+y)}{\sin(x-y)}=\frac{a+b}{a-b}$
by componendo and dividendo we get
$\large\frac{\sin(x+y)+\sin(x-y)}{\sin(x-y)-\sin(x-y)}=\frac{a+b+a-b}{a+b-a+b}$
$\large\frac{\sin(x+y)+\sin(x-y)}{\sin(x-y)-\sin(x-y)}=\frac{a}{b}$
$\Rightarrow \large\frac{2\sin\Large\frac{x+y+x-y}{2}.\cos \Large\frac{x+y-x+y}{2}}{2\cos\Large\frac{x+y+x-y}{2}.\sin \Large\frac{x+y-x+y}{2}}=\frac{a}{b}$
$\Rightarrow \large\frac{\sin x.\cos y}{\cos x.\sin y}=\frac{a}{b}$
$\Rightarrow \tan x.\cot y=\large\frac{a}{b}$
$\Rightarrow \large\frac{\tan x}{\tan y}=\frac{a}{b}$
Hence proved.
answered Apr 18, 2014 by sreemathi.v
 

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