Question

Given that $\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)$, then $\tan \left( {\dfrac{\alpha }{2}} \right)\tan \left( {\dfrac{\beta }{2}} \right)$ is equal to:
A. $\dfrac{1}{2}$
B. $\dfrac{1}{3}$
C. $\dfrac{1}{4}$
D. $\dfrac{1}{8}$

Answer 1

The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $\cos C - \cos D = - 2\sin \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right)$ and $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem. We must know the transposition rule for simplification of the equation by shifting the terms from one side of the equation to another.

Complete step by step answer:
In the given problem, we are given that: $\cos \left( {\dfrac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)$.
Now, we subtract $\cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)$ from both sides of the equation. So, we get,
$\Rightarrow \cos \left( {\dfrac{{\alpha - \beta }}{2}} \right) - \cos \left( {\dfrac{{\alpha + \beta }}{2}} \right) = \cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)$
Now, we must know the formula for difference of cosines for two angles $\cos C - \cos D = - 2\sin \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right)$. So, using this trigonometric formula, we get,
$\Rightarrow - 2\sin \left[ {\dfrac{{\left( {\dfrac{{\alpha - \beta }}{2} + \dfrac{{\alpha + \beta }}{2}} \right)}}{2}} \right]\sin \left[ {\dfrac{{\left( {\dfrac{{\alpha - \beta }}{2} - \dfrac{{\alpha + \beta }}{2}} \right)}}{2}} \right] = \cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)$
$\Rightarrow - 2\sin \left[ {\dfrac{\alpha }{2}} \right]\sin \left[ { - \dfrac{\beta }{2}} \right] = \cos \left( {\dfrac{{\alpha + \beta }}{2}} \right)$

Now, expanding the right side of the equation using the compound angle formula for cosine $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$, we get,
$\Rightarrow - 2\sin \left[ {\dfrac{\alpha }{2}} \right]\sin \left[ { - \dfrac{\beta }{2}} \right] = \cos \left( {\dfrac{\alpha }{2}} \right)\cos \left( {\dfrac{\beta }{2}} \right) - \sin \left( {\dfrac{\alpha }{2}} \right)\sin \left( {\dfrac{\beta }{2}} \right)$
We know that sine is an odd function. So, $\sin \left( { - x} \right) = - \sin x$.
Now, dividing both sides of the equation by $\cos \left( {\dfrac{\alpha }{2}} \right)\cos \left( {\dfrac{\beta }{2}} \right)$, we get,
$\Rightarrow 2\dfrac{{\sin \left( {\dfrac{\alpha }{2}} \right)\sin \left( {\dfrac{\beta }{2}} \right)}}{{\cos \left( {\dfrac{\alpha }{2}} \right)\cos \left( {\dfrac{\beta }{2}} \right)}} = \dfrac{{\cos \left( {\dfrac{\alpha }{2}} \right)\cos \left( {\dfrac{\beta }{2}} \right) - \sin \left( {\dfrac{\alpha }{2}} \right)\sin \left( {\dfrac{\beta }{2}} \right)}}{{\cos \left( {\dfrac{\alpha }{2}} \right)\cos \left( {\dfrac{\beta }{2}} \right)}}$

Now, simplifying the expression, we get,
$\Rightarrow 2\tan \left( {\dfrac{\alpha }{2}} \right)\tan \left( {\dfrac{\beta }{2}} \right) = 1 - \tan \left( {\dfrac{\alpha }{2}} \right)\tan \left( {\dfrac{\beta }{2}} \right)$
Transposing all the constant terms to right side of the equation and trigonometric functions to left side of the equation, we get,
$\Rightarrow 2\tan \left( {\dfrac{\alpha }{2}} \right)\tan \left( {\dfrac{\beta }{2}} \right) + \tan \left( {\dfrac{\alpha }{2}} \right)\tan \left( {\dfrac{\beta }{2}} \right) = 1$
$\Rightarrow 3\tan \left( {\dfrac{\alpha }{2}} \right)\tan \left( {\dfrac{\beta }{2}} \right) = 1$
Dividing both sides by three, we get,
$\therefore \tan \left( {\dfrac{\alpha }{2}} \right)\tan \left( {\dfrac{\beta }{2}} \right) = \dfrac{1}{3}$
Hence, the value of $\tan \left( {\dfrac{\alpha }{2}} \right)\tan \left( {\dfrac{\beta }{2}} \right)$ is $\dfrac{1}{3}$.

Hence, option B is the correct answer.

Note: We must have a strong grip over the concepts of trigonometry, related formulae and rules to ace these types of questions. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such types of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers.