Question

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

Answer 1

We have,

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

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

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

$\sin \alpha =a(\sin \alpha \cos \beta +\sin \beta \cos \alpha )$

$\sin \alpha =a\sin (\alpha +\beta )$ …….. (1)

Similarly,

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

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

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

$\sin \beta =b(\sin \alpha \cos \beta +\cos \alpha \sin \beta )$

$\sin \beta =b\sin (\alpha +\beta )$ ……. (2)

From equation (1) and (2), we get

$\frac{\sin \alpha }{\sin \beta }=\frac{a}{b}$

R.H.S

Hence, proved.