Question

Let $\alpha$ and $\beta$ are 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)

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

$2(\cos \beta -\cos \alpha )+\cos \alpha \cos \beta =1$

$2(\cos \beta -\cos \alpha )=1-\cos \alpha \cos \beta$

$(\cos \beta -\cos \alpha )=\frac{1-\cos \alpha \cos \beta }{2}$

$cos2x=\frac{1-{\tan }^{2}x}{1+ta{n}^{2}x}$

let,$\tan \frac{\alpha }{2}=x,\tan \frac{\beta }{2}=y$

$\therefore cos\alpha =\frac{1-x^2}{1+x^2}$ and $cos\beta =\frac{1-y^2}{1+y^2}$

$\Rightarrow 2[\frac{1-y^2}{1+y^2}-\frac{1-x^2}{1+x^2}]=1-\frac{(1-x^2)(1-y^2)}{(1+x^2)(1+y^2)}\Rightarrow 2\left[\right(1+x^2\left)\right(1-y^2)-({-x}^2\left)\right(1+y^2\left)\right]=(1+x^2)(1+y^2)-(1-x^2)(1-y^2)\Rightarrow 4(x^2-y^2)=2(x^2+y^2)\Rightarrow x^2=3y^2\Rightarrow x=\pm \sqrt{3}y\Rightarrow \tan \frac{\alpha }{2}\pm \sqrt{3}\tan \frac{\beta }{2}=0$