Question

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

Answer 1

$\tan \beta =\frac{\tan \alpha +\tan y}{1+\tan \alpha \tan y}$

$=\frac{\frac{\sin \alpha }{\cos \alpha }+\frac{\sin y}{\cos y}}{1+\frac{\sin \alpha }{\cos \alpha }\cdot \frac{\sin y}{\cos y}}$

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

$\text{Now}\sin 2\beta =\frac{2\tan \beta }{1+{\tan }^2\beta }$

$=2\left[\frac{\frac{\sin (\alpha +y)}{\cos (\alpha -y)}}{1+\frac{{\sin }^2(\alpha +y)}{{\cos }^2(\alpha -y)}}\right]$

$=2\left[\frac{\sin (\alpha +y)\cos (\alpha -y)}{(\cos \alpha \cos y+\sin \alpha \sin y{)}^2+(\sin \alpha \cos y+\cos \alpha \sin y{)}^2}\right]$

$=2\left[\frac{(\sin \alpha \cos y+\cos \alpha \sin y)(\cos \alpha \cos y+\sin \alpha \sin y)}{\begin{array}{c}{\cos }^2\alpha {\cos }^{2}y+{\sin }^2\alpha {\sin }^{2}y+2\sin \alpha \cos \alpha \cos y\sin y\\ +{\sin }^2\alpha {\cos }^{2}y+{\cos }^2\alpha {\sin }^{2}y+2\cos \alpha \cos y\sin \alpha \sin y\end{array}}\right]$

$=2\left[\frac{\sin \alpha \cos \alpha {\cos }^{2}y+{\sin }^2\alpha \cos y\sin y+{\cos }^2\alpha \sin y\cos y+{\sin }^{2}y\cos \alpha \sin \alpha }{{\cos }^{2}y+{\sin }^{2}y+4\sin \alpha \cos \alpha \cos y\sin y}\right]$

$=2\left[\frac{\sin \alpha \cos \alpha +\sin y\cos y}{1+\sin 2\alpha \sin 2y}\right]$

$=\frac{\sin 2\alpha +\sin 2y}{1+\sin 2\alpha \sin 2y}=RHS.$

Hence LHS=RHS proof