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

Answer 1

Answer: (C), (D)

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

Substitute the values of $\cos \alpha$ and $\cos \beta$ in terms of the tangents of their half angles.

$\cos \alpha =\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}$ and $\cos \beta =\frac{1-{\tan }^2\frac{\beta }{2}}{1+{\tan }^2\frac{\beta }{2}}$

Let ${\tan }^2\frac{\alpha }{2}=a$ and ${\tan }^2\frac{\beta }{2}=b$

Substituting the above values in the expression,

$2(\frac{1-b}{1+b}-\frac{1-a}{1+a})+\left(\frac{1-a}{1+a}\right)\left(\frac{1-b}{1+b}\right)=1$

$\Rightarrow 2\left(\frac{1-b+a-ab-1+a-b+ab}{1+a+b+ab}\right)+\frac{1-a-b+ab}{1+a+b+ab}=1$

$\Rightarrow 4(a-b)+1-a-b+ab=1+a+b+ab$

$\Rightarrow 4a-4b=2b+2a$

$\Rightarrow 2a=6b\Rightarrow a=3b$

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

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

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

$\therefore \tan \frac{\alpha }{2}+\sqrt{3}\tan \frac{\beta }{2}=0$ or $\tan \frac{\alpha }{2}-\sqrt{3}\tan \frac{\beta }{2}=0$

Hence options C and D are correct.