Question

If $\alpha +\beta =\gamma$ and $\frac{tan\alpha }{tan\beta }=\frac{x}{y},$ the prove that sin $(\alpha -\beta )=\frac{x-y}{x+y}sin\gamma$.

Answer 1

We have,

$\alpha +\beta =\chi$ ……. (1)

$\frac{\tan \alpha }{\tan \beta }=\frac{x}{y}$

Using componendo and divideno rule,

$\frac{\tan \alpha +\tan \beta }{\tan \alpha -\tan \beta }=\frac{x+y}{x-y}$

$\frac{\frac{\sin \alpha }{\cos \alpha }+\frac{\sin \beta }{\cos \beta }}{\frac{\sin \alpha }{\cos \alpha }-\frac{\sin \beta }{\cos \beta }}=\frac{x+y}{x-y}$

$\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\sin \alpha \cos \beta -\cos \alpha \sin \beta }=\frac{x+y}{x-y}$

$\frac{\sin (\alpha +\beta )}{\sin (\alpha -\beta )}=\frac{x+y}{x-y}$

$\text{For}\text{recipocal}$

$\frac{\sin (\alpha -\beta )}{\sin (\alpha +\beta )}=\frac{x-y}{x+y}$

$\text{From}\text{equation}\left(\text{1}\right)$

$\frac{\sin (\alpha -\beta )}{\sin \chi }=\frac{x-y}{x+y}$

$\sin (\alpha -\beta )=\frac{x-y}{x+y}\sin \chi$

$\text{Hence}\text{proved}\text{.}$