Question

Let $\alpha $ and $\beta $ be two non-zero real numbers such that $2\left( \cos \beta -\cos \alpha \right)+\cos \alpha \cos \beta =1$. Then which of the following is/are correct?
(a) $\sqrt{3}\tan \left( \dfrac{\alpha }{2} \right)+\tan \left( \dfrac{\beta }{2} \right)=0$
(b) $\sqrt{3}\tan \left( \dfrac{\alpha }{2} \right)-\tan \left( \dfrac{\beta }{2} \right)=0$
(c) $\tan \left( \dfrac{\alpha }{2} \right)+\sqrt{3}\tan \left( \dfrac{\beta }{2} \right)=0$
(d) $\tan \left( \dfrac{\alpha }{2} \right)-\sqrt{3}\tan \left( \dfrac{\beta }{2} \right)=0$

Answer 1

Hint: First, before proceeding for this, we must know the following trigonometric formulas to get the answer easily as $\cos \alpha =\dfrac{1-{{\tan }^{2}}\dfrac{\alpha }{2}}{1+{{\tan }^{2}}\dfrac{\alpha }{2}}$ and $\cos \beta =\dfrac{1-{{\tan }^{2}}\dfrac{\beta }{2}}{1+{{\tan }^{2}}\dfrac{\beta }{2}}$. Then, we can see that the values above are bit large and the expression becomes complex after substitution, so let us suppose as ${{\tan }^{2}}\dfrac{\alpha }{2}=a$ and ${{\tan }^{2}}\dfrac{\beta }{2}=b$. Then, by replacing the value a and b with their original values, we get the relation, we get the final result.

Complete step-by-step answer:
In this question, we are supposed to find the correct option/s when the given condition is $\alpha $and $\beta $be two non-zero real numbers such that $2\left( \cos \beta -\cos \alpha \right)+\cos \alpha \cos \beta =1$.
So, before proceeding for this, we must know the following trigonometric formulas to get the answer easily as:
$\cos \alpha =\dfrac{1-{{\tan }^{2}}\dfrac{\alpha }{2}}{1+{{\tan }^{2}}\dfrac{\alpha }{2}}$ and $\cos \beta =\dfrac{1-{{\tan }^{2}}\dfrac{\beta }{2}}{1+{{\tan }^{2}}\dfrac{\beta }{2}}$
Now, we can see that the values above are bit large and the expression becomes complex after substitution, so let us suppose as:
${{\tan }^{2}}\dfrac{\alpha }{2}=a$ and ${{\tan }^{2}}\dfrac{\beta }{2}=b$
Then, after substituting the value in the given expression, we get:
$2\left( \dfrac{1-b}{1+b}-\dfrac{1-a}{1+a} \right)+\left( \dfrac{1-a}{1+a} \right)\left( \dfrac{1-b}{1+b} \right)=1$
Then, by solving the above expression, we get:
$\begin{align}
  & 2\left( \dfrac{\left( 1-b \right)\left( 1+a \right)-\left( 1-a \right)\left( 1+b \right)}{\left( 1+a \right)\left( 1+b \right)} \right)+\left( \dfrac{1-a}{1+a} \right)\left( \dfrac{1-b}{1+b} \right)=1 \\
 & \Rightarrow 2\left( \dfrac{1-b+a-ab-1+a-b+ab}{1+a+b+ab} \right)+\dfrac{1-a-b+ab}{1+a+b+ab}=1 \\
 & \Rightarrow 4\left( a-b \right)+1-a-b+ab=1+a+b+ab \\
 & \Rightarrow 4a-4b=2b+2a \\
 & \Rightarrow 2a=6b \\
 & \Rightarrow a=3b \\
\end{align}$
Then, by replacing the value a and b with their original values, we get the relation as:
$\begin{align}
  & {{\tan }^{2}}\dfrac{\alpha }{2}=3{{\tan }^{2}}\dfrac{\beta }{2} \\
 & \Rightarrow {{\tan }^{2}}\dfrac{\alpha }{2}-3{{\tan }^{2}}\dfrac{\beta }{2}=0 \\
\end{align}$
Now, by solving further, we get:
$\left( \tan \dfrac{\alpha }{2}+\sqrt{3}\tan \dfrac{\beta }{2} \right)\left( \tan \dfrac{\alpha }{2}-\sqrt{3}\tan \dfrac{\beta }{2} \right)=0$
So, we get the two answers possible from the above conditions as:
$\left( \tan \dfrac{\alpha }{2}+\sqrt{3}\tan \dfrac{\beta }{2} \right)=0$ and $\left( \tan \dfrac{\alpha }{2}-\sqrt{3}\tan \dfrac{\beta }{2} \right)=0$
Hence, options (c) and (d) are correct.

Note: Now, to solve these types of questions we need to know some of the basic conversions beforehand so that we can easily proceed in these types of questions. Then, some of the basic conversions are:
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Moreover, we must be very careful with trigonometric formulas substitution so that we will not proceed in a wrong way.