JEE Advance - Mathematics (2017 - Paper 2 Offline - No. 11)
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?
$$\sqrt 3 \tan \left( {{\alpha \over 2}} \right) - \tan \left( {{\beta \over 2}} \right) = 2$$
$$\tan \left( {{\alpha \over 2}} \right) - \sqrt 3 \tan \left( {{\beta \over 2}} \right) = 0$$
$$\tan \left( {{\alpha \over 2}} \right) + \sqrt 3 \tan \left( {{\beta \over 2}} \right) = 0$$
$$\sqrt 3 \tan \left( {{\alpha \over 2}} \right) + \tan \left( {{\beta \over 2}} \right) = 2$$
Explanation
$$2(\cos \beta - \cos \alpha ) + \cos \alpha \cos \beta = 1$$
or $$4(\cos \beta - \cos \alpha ) + 2\cos \alpha \cos \beta = 2$$
$$ \Rightarrow 1 - \cos \alpha + \cos \beta - \cos \alpha \cos \beta $$
$$ = 3 + 3\cos \alpha - 3\cos \beta - 3\cos \alpha \cos \beta $$
$$ \Rightarrow (1 - \cos \alpha )(1 + \cos \beta )$$
$$ = 3(1 + \cos \alpha )(1 - \cos \beta )$$
$$ \Rightarrow {{(1 - \cos \alpha )} \over {(1 + \cos \alpha )}} = {{3(1 - \cos \beta )} \over {1 + \cos \beta }}$$
$$ \Rightarrow {\tan ^2}{\alpha \over 2} = 3{\tan ^2}{\beta \over 2}$$
$$ \therefore $$ $$\tan {\alpha \over 2} \pm \sqrt 3 \tan {\beta \over 2} = 0$$
or $$4(\cos \beta - \cos \alpha ) + 2\cos \alpha \cos \beta = 2$$
$$ \Rightarrow 1 - \cos \alpha + \cos \beta - \cos \alpha \cos \beta $$
$$ = 3 + 3\cos \alpha - 3\cos \beta - 3\cos \alpha \cos \beta $$
$$ \Rightarrow (1 - \cos \alpha )(1 + \cos \beta )$$
$$ = 3(1 + \cos \alpha )(1 - \cos \beta )$$
$$ \Rightarrow {{(1 - \cos \alpha )} \over {(1 + \cos \alpha )}} = {{3(1 - \cos \beta )} \over {1 + \cos \beta }}$$
$$ \Rightarrow {\tan ^2}{\alpha \over 2} = 3{\tan ^2}{\beta \over 2}$$
$$ \therefore $$ $$\tan {\alpha \over 2} \pm \sqrt 3 \tan {\beta \over 2} = 0$$
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