Question

If $\cos \theta =\frac{\cos \alpha -\cos \beta }{1-\cos \alpha \cos \beta }$, then one of the values of $\tan \frac{\theta }{2}$ is :

• A. $\tan \frac{\theta }{2}\cot \frac{\beta }{2}$
• B. $\tan \frac{\beta }{2}\cot \frac{\alpha }{2}$
• C. $\sin \frac{\alpha }{2}\sin \frac{\beta }{2}$
• D. None of these

Answer 1

Answer: (A)

${\tan }^2\frac{\theta }{2}=\frac{1-\cos \theta }{1+\cos \theta }=\frac{1-\frac{\cos \alpha -\cos \beta }{1-\cos \alpha \cos \beta }}{1+\frac{\cos \alpha -\cos \beta }{1-\cos \alpha \cos \beta }}$
$=\frac{1-\cos \alpha \cos \beta -\cos \alpha +\cos \beta }{1-\cos \alpha \cos \beta +\cos \alpha -\cos \beta }$
$=\frac{(1-\cos \alpha )+\cos \beta (1-\cos \alpha )}{(1+\cos \alpha )-\cos \beta (1+\cos \alpha )}$
$=\frac{(1-\cos \alpha )+(1+\cos \beta )}{(1+\cos \alpha )(1-\cos \beta )}={\tan }^2\frac{\alpha }{2}{\cot }^2\frac{\beta }{2}$
$\therefore \tan \frac{\theta }{2}=\pm \tan \frac{\alpha }{2}\cot \frac{\beta }{2}$