Question

If $\cos \theta =\frac{\cos \alpha -\cos \beta }{1-\cos \alpha \cos \beta }$ , then $\tan \frac{\theta }{2}$ is equal to

• A. $\pm \tan \left(\frac{\alpha }{2}\right)\tan \left(\frac{\beta }{2}\right)$
• B. $\pm \tan \left(\frac{\alpha }{2}\right)\cot \left(\frac{\beta }{2}\right)$
• C. $\pm \tan \left(\frac{\beta }{2}\right)\cot \left(\frac{\alpha }{2}\right)$
• D. None of these

Answer 1

The correct option is B $\pm \tan \left(\frac{\alpha }{2}\right)\cot \left(\frac{\beta }{2}\right)$

${\tan }^2\frac{\theta }{2}=\frac{{(1-\cos \theta )}^2}{{\sin }^2\theta }$

$=\frac{{(1-\cos \theta )}^2}{1-{\cos }^2\theta }$

$=\frac{1-\cos \theta }{1+\cos \theta }$

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

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

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

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

$=\frac{2{\sin }^2\frac{\alpha }{2}{2\cos }^2\frac{\beta }{2}}{2{\cos }^2\frac{\alpha }{2}{2\sin }^2\frac{\beta }{2}}$

$={\tan }^2\frac{\alpha }{2}{\cot }^2\frac{\beta }{2}$

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