Question

If $\cos \theta =\frac{\cos \alpha \cos \beta }{1-\sin \alpha \sin \beta }$ then $\tan \frac{\theta }{2}=\frac{\tan \frac{\alpha }{2}-\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}$.

• A. True
• B. False

Answer 1

The correct option is A True

Given

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

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

Applying Compoundo and Dividendo Rule.

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

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

${\tan }^2\frac{\theta }{2}=\frac{{\sin }^2\left(\frac{\alpha -\beta }{2}\right)}{{\cos }^2\left(\frac{\alpha +\beta }{2}\right)}$

$\tan \frac{\theta }{2}=\frac{\sin \left(\frac{\alpha -\beta }{2}\right)}{\cos \left(\frac{\alpha +\beta }{2}\right)}$

$\tan \frac{\theta }{2}=\frac{\sin \frac{\alpha }{2}\cos \frac{\beta }{2}-\sin \frac{\beta }{2}\cos \frac{\alpha }{2}}{\cos \frac{\alpha }{2}\cos \frac{\beta }{2}-\sin \frac{\alpha }{2}\sin \frac{\beta }{2}}$

$\tan \frac{\theta }{2}=\frac{\tan \frac{\alpha }{2}-\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}$