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?

The correct options are
A $\tan \left(\frac{\alpha }{2}\right)+\sqrt{3}\tan \left(\frac{\beta }{2}\right)=0$
C $\tan \left(\frac{\alpha }{2}\right)-\sqrt{3}\tan \left(\frac{\beta }{2}\right)=0$
$2(\cos \beta -\cos \alpha )+\cos \alpha \cos \beta =1\cdots \left(1\right)$

Since, $\cos 2\alpha =\frac{1-{\tan }^2\alpha }{1-{\tan }^2\alpha }$

i.e.,
$\cos \alpha =\frac{1-a}{1+a};\text{where}a={\tan }^2\frac{\alpha }{2}$
$\cos \beta =\frac{1-b}{1+b};\text{where}b={\tan }^2\frac{\beta }{2}$

From $\left(1\right),$ we get
$2[\left(\frac{1-b}{1+b}\right)-\left(\frac{1-a}{1+a}\right)]+[\left(\frac{1-a}{1+a}\right).\left(\frac{1-b}{1+b}\right)]-1=0$

$\Rightarrow \frac{2(1-b)(1+a)-2(1-a)(1+b)+(1-a)(1-b)-(1+a)(1+b)}{(1+a)(1+b)}=0$

$\Rightarrow 2(1-b)(1+a)-2(1-a)(1+b)+(1-a)(1-b)-(1+a)(1+b)=0$
$\Rightarrow a=3b$

$\Rightarrow {\tan }^2\frac{\alpha }{2}=3{\tan }^2\frac{\beta }{2}$
$\Rightarrow {\tan }^2\frac{\alpha }{2}-\sqrt{3}{\tan }^2\frac{\beta }{2}=0$

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

$\Rightarrow \tan \frac{\alpha }{2}-\sqrt{3}\tan \frac{\beta }{2}=0,\text{and}\tan \frac{\alpha }{2}+\sqrt{3}\tan \frac{\beta }{2}=0$
Therefore, Option (1) and (3) are correct.