Question

If $\tan \beta =\frac{\tan \alpha +\tan \beta }{1+\tan \alpha \tan \beta }$, then $\sin c\beta =\frac{\sin 2\alpha +\sin 2\gamma }{1+\sin 2\alpha \sin 2\gamma }$ Find $c$.

Answer 1

$\tan \beta =\frac{\frac{\sin \alpha }{\cos \alpha }+\frac{\sin \gamma }{\cos \gamma }}{1+\frac{\sin \alpha }{\cos \alpha }\frac{\sin \gamma }{\cos \gamma }}=\frac{\sin (\alpha +\gamma )}{\cos (\alpha -\gamma )}$

$\sin 2\beta =\frac{2\tan \beta }{1+{\tan }^2\beta }=\frac{2\frac{\sin (\alpha +\gamma )}{\cos (\alpha -\gamma )}}{1+\frac{{\sin }^2(\alpha +\gamma )}{{\cos }^2(\alpha -\gamma )}}$

$=\frac{2\sin (\alpha +\gamma )\cos (\alpha -\gamma )}{{\cos }^2(\alpha -\gamma )+{\sin }^2(\alpha +\gamma )}$

$=\frac{\sin 2\alpha +\sin 2\gamma }{\frac{1+\cos 2(\alpha -\gamma )}{2}+\frac{1-\cos 2(\alpha +\gamma )}{2}}$

$=\frac{\sin 2\alpha +\sin 2\gamma }{1+\frac{1}{2}\times 2\sin 2\alpha \sin \gamma }=\frac{\sin 2\alpha +\sin 2\gamma }{1+\sin 2\alpha \sin 2\gamma }$
Ans: c = 2