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)

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