Question

If $\cos \theta =\frac{\cos \alpha -\cos \beta }{1-\cos \alpha \cos \beta }$, then show that $\tan \frac{\theta }{2}$ $=\pm \tan \frac{\alpha }{2}$ $\cot \frac{\beta }{2}$

Answer 1

$\cos \theta =\frac{\cos \alpha -\cos \beta }{1-\cos \alpha \cos \beta }$

or, $\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}=\frac{\cos \alpha -\cos \beta }{1-\cos \alpha \cos \beta }$

or, $\frac{1+{\tan }^2\frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}=\frac{1-\cos \alpha \cos \beta }{\cos \alpha -\cos \beta }$

or, ${\tan }^2\frac{\theta }{2}=\frac{1-\cos \alpha \cos \beta -\cos \alpha +\cos \beta }{1-\cos \alpha \cos \beta +\cos \alpha -\cos \beta }$

or, ${\tan }^2\frac{\theta }{2}=\frac{(1-\cos \alpha )(1+\cos \beta )}{(1+\cos \alpha )(1-\cos \beta )}$

or, ${\tan }^2\frac{\theta }{2}={\tan }^2\frac{\alpha }{2}{\cos }^2\frac{\beta }{2}$

or, $\tan \frac{\theta }{2}=\pm \tan \frac{\alpha }{2}\cot \frac{\beta }{2}$