Family of lines $\sin\alpha x +\sin\beta y +\sin\gamma =0$

Question

Problem :

If $\sin(\alpha + \beta)\sin(\alpha -\beta) =\sin\gamma (2\sin\beta +\sin\gamma), 0 < \alpha , \beta ,\gamma <\pi$ then the family of lines $\sin\alpha x +\sin\beta y +\sin\gamma =0$ passes through

(a) $(-1,1)$

(b) $(1,1)$

(c) $(1,-1)$

(d) $(-1,-1)$

My approach :

$\sin(\alpha + \beta)\sin(\alpha -\beta) =\sin\gamma (2\sin\beta +\sin\gamma) $

$\Rightarrow \sin^2\alpha -\sin^2\beta = \sin\gamma (2\sin\beta +\sin\gamma) $

By using $\sin(A+B) \sin(A-B) = \sin^2A -\sin^2B$

But I am not getting any clue how to proceed further.. please guide thanks..

Answer 1

On rearrangement, We have $$\sin^2\alpha=\sin^2\beta+2\sin\beta\sin\gamma+\sin^2\gamma=(\sin\beta+\sin\gamma)^2$$

As $0<\alpha,\beta,\gamma<\pi,$ all the sine ratios are $>0$

$$\implies\sin\alpha=\sin\beta+\sin\gamma$$

Now use $\displaystyle\sin(-y)=-\sin y$