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,

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

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

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

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

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

Similarly,

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

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

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

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

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

From equation $\left(1\right)$ and $\left(2\right)$, we get

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

Hence, proved.