Let $\alpha$ and $\beta$ be nonzero real numbers such that $2 (\cos \alpha-\cos\beta)+\cos\alpha\cos\beta=1$. Then which of the following is/are true ?

a) $\sqrt3\tan(\frac\alpha2)+\tan(\frac\beta2)=0$

b) $\sqrt3\tan(\frac\alpha2)-\tan(\frac\beta2)=0$

c) $\tan(\frac\alpha2)+\sqrt3\tan(\frac\beta2)=0$

d) $\tan(\frac\alpha2)-\sqrt3\tan(\frac\beta2)=0$

I solved the equation by converting all cosines to half angles and the answers options c) and d), seems pretty straightforward from there. But, when the question was asked in an engineering entrance exam (JEE ADVANCED 2017), the question was given as BONUS that would mean either the question was not complete, was wrong or might be some other mistake in understanding the question. I do not understand what was the problem. Can someone help me point out the ambiguity or mistake in the question or give a solution that differs from mine.


Firstly, the question doesn’t specify the ranges of $\alpha,\beta$. The equation given in the question and any one of the options can be simultaneously true for some values of $\alpha, \beta$. (c) and (d) just happens to be true for intervals of non-zero length while (a) and (b) are true for some discrete values. If the question wants to guide us to answer (c) or (d), “$\forall\alpha$ in $[\cdot,\cdot]$ and $\forall\beta$ in $[\cdot,\cdot]$” should be added.

Secondly, none of the options is true if $\alpha,\beta$ are odd multiples of $\pi$, where $\tan$ is undefined.

Thirdly, (c) and (d) cannot be simultaneously true unless $\alpha,\beta$ are even multiples of $\pi$. Without specifying the ranges of $\alpha,\beta$, whether (c) or (d) is true cannot be determined.


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

So $\displaystyle \cos \alpha(\cos \beta+2)=1+2\cos \beta \Rightarrow \cos \alpha =\frac{2\cos \beta+1}{2+\cos \beta}$

So $\displaystyle \frac{1}{\cos \alpha} = \frac{2+\cos \beta}{2\cos \beta +1}$

Using Componendo Dividedno, We have

$\displaystyle \frac{1+\cos \alpha}{1-\cos \alpha} = \frac{2+\cos \beta+2\cos \beta+1}{(2+\cos \beta)-(2\cos \beta +1)} = \frac{3(\cos \beta+1)}{1-\cos \beta}$

So $\displaystyle \frac{2\sin^2\frac{\alpha}{2}}{2\cos^2 \frac{\alpha}{2}} = 3\cdot \frac{2\sin^2\frac{\beta}{2}}{2\cos^2 \frac{\beta}{2}}$

So we have $$\tan \frac{\alpha}{2} = \pm\sqrt{3}\tan \frac{\beta}{2}.$$

  • $\begingroup$ Thank you durgesh ji for the solution. Your answer matches mine. But I asked for ambiguity if any in the question because of which the IITs gave this question as BONUS. $\endgroup$
    – sonu
    Jun 5 '18 at 12:25

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