Question

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

• A. ${\tan }^2\frac{\alpha }{2}$
• B. ${\tan }^2\frac{\beta }{2}$
• C. ${\tan }^2\frac{\theta }{2}$
• D. ${\cot }^2\frac{\beta }{2}$

Answer 1

The correct option is B ${\tan }^2\frac{\beta }{2}$

$\cos \theta =\cos \alpha \cos \beta$

$\Rightarrow \frac{\cos \theta }{\cos \alpha }=\cos \beta$

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

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

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

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

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