Question

Let $\alpha$ and $\beta$ be nonzero real numbers such that $2(\cos \beta -\cos \alpha )+\cos \alpha \cos \beta =1$. Then which of the following is/are true?

• A. $\tan \left(\frac{\alpha }{2}\right)+\sqrt{3}\tan \left(\frac{\beta }{2}\right)=0$
• B. $\sqrt{3}\tan \left(\frac{\alpha }{2}\right)+\tan \left(\frac{\beta }{2}\right)=0$
• C. $\tan \left(\frac{\alpha }{2}\right)-\sqrt{3}\tan \left(\frac{\beta }{2}\right)=0$
• D. $\sqrt{3}\tan \left(\frac{\alpha }{2}\right)-\tan \left(\frac{\beta }{2}\right)=0$

Answer 1

Answer: (A), (C)

$\cos \beta (2+\cos \alpha )=1+2\cos \alpha$
$\frac{\cos \beta }{1}=\frac{1+2\cos \alpha }{2+\cos \alpha }$
$\frac{\cos \beta +1}{\cos \beta -1}=\frac{3(\cos \alpha +1)}{\cos \alpha -1}$
$\frac{2{\cos }^2\frac{\beta }{2}}{-2{\sin }^2\frac{\beta }{2}}=\frac{3\times 2{\cos }^2\frac{\alpha }{2}}{-2{\sin }^2\frac{\alpha }{2}}$
${\tan }^2\frac{\alpha }{2}=3{\tan }^2\frac{\beta }{2}$
$\Rightarrow \tan \frac{\alpha }{2}=\pm \sqrt{3}\tan \frac{\beta }{2}$