Question

Let $\alpha$ and $\beta$ be non-zero 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)

$\tan \left(\frac{\alpha }{2}\right)+\sqrt{3}\tan \left(\frac{\beta }{2}\right)=0$

As $2(\cos \beta –\cos \alpha )=1–\cos \alpha .\cos \beta$
$\Rightarrow \cos \alpha =\frac{2\cos \beta -1}{2-\cos \beta }$
Using componendo and dividendo, $\Rightarrow \frac{1-\cos \alpha }{1+\cos \alpha }=3\left(\frac{1-\cos \beta }{1+\cos \beta }\right)$
$\Rightarrow {\tan }^2\frac{\alpha }{2}-3{\tan }^2\frac{\beta }{2}=0$
So, $\tan \frac{\alpha }{2}+\sqrt{3}\tan \frac{\beta }{2}=0$
or $\tan \frac{\alpha }{2}-\sqrt{3}\tan \frac{\beta }{2}=0$