Question

$\tan \alpha =\frac{a\sin \beta }{1-a\cos \beta }$ and $\tan \beta =\frac{b\sin \alpha }{1-b\cos \alpha }$ then show that $\frac{\sin \alpha }{\sin \beta }=\frac{a}{b}$

Answer 1

$\tan \alpha =\frac{a\sin \beta }{1-a\cos \beta }Given$

$\Rightarrow \frac{\sin \alpha }{\cos \alpha }=\frac{a\sin \beta }{1-a\cos \beta }$

$\Rightarrow \sin \alpha -a\cos \beta \sin \alpha =a\sin \beta \cos \alpha$

$\Rightarrow \sin \alpha =a\sin (\beta +\alpha )\left(1\right)$

$Also\tan \beta =\frac{b\sin \alpha }{1-b\cos \alpha }$

$\Rightarrow \frac{\sin \beta }{\cos \beta }=\frac{b\sin \alpha }{1-b\cos \alpha }$

$\Rightarrow \sin \beta -b\cos \alpha \sin \beta =b\sin \alpha \cos \beta$

$\Rightarrow \sin \beta =b\sin (\alpha +\beta )\left(2\right)$

$Lhs=\frac{\sin \alpha }{\sin \beta }=\frac{a\sin (\alpha +\beta )}{b\sin (\alpha +\beta )}\left[\text{by}\right(1\left)\&\right(2\left)\right]$

$=\frac{a}{b}=Rhs$

Hence proof