Question

If $\dfrac{{\cos A}}{{\cos B}} = \dfrac{{\sin \left( {C - \theta } \right)}}{{\sin \left( {C + \theta } \right)}}$, then $\tan \theta $ is equal to?
(A) $\tan \left( {\dfrac{{A + B}}{2}} \right)\tan \left( {\dfrac{{A - B}}{2}} \right)\tan \dfrac{C}{2}$
(B) $\tan \left( {\dfrac{{A + B}}{2}} \right)\tan \left( {\dfrac{{A - B}}{2}} \right)\tan C$
(C) $\tan \left( {\dfrac{{A + B}}{2}} \right)\tan \left( {\dfrac{{A - B}}{2}} \right)\sin \dfrac{C}{2}$
(D) $\tan \left( {\dfrac{{A + B}}{2}} \right)\tan \left( {\dfrac{{A - B}}{2}} \right)\cos \dfrac{C}{2}$

Answer 1

Hint: Start with the right hand side of the equation and apply formulas \[\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\] and \[\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B\]. Then use componendo and dividendo after simplification. After that use trigonometric formulas of \[\cos C + \cos D\] and \[\cos C - \cos D\] to convert the expression in the form of $\tan A,{\text{ }}\tan B{\text{ and }}\tan \theta $

Complete step-by-step solution:
According to the question, we have been given a trigonometric equation which is:
$ \Rightarrow \dfrac{{\cos A}}{{\cos B}} = \dfrac{{\sin \left( {C - \theta } \right)}}{{\sin \left( {C + \theta } \right)}}$
From this equation, we have to find out the value of $\tan \theta $ in terms of other variables.
Given below are two trigonometric formulas that we are going to use here:
\[
   \Rightarrow \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B \\
   \Rightarrow \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B \\
 \]
If we use these two formulas on the right hand side of the above equation, we’ll get:
$ \Rightarrow \dfrac{{\cos A}}{{\cos B}} = \dfrac{{\sin C\cos \theta - \cos C\sin \theta }}{{\sin C\cos \theta + \cos C\sin \theta }}$
If two ratios are equal, then we can apply componendo and dividend on them. Applying this theorem for the above equation, we’ll get:
$ \Rightarrow \dfrac{{\cos A + \cos B}}{{\cos A - \cos B}} = \dfrac{{\sin C\cos \theta - \cos C\sin \theta + \sin C\cos \theta + \cos C\sin \theta }}{{\sin C\cos \theta - \cos C\sin \theta - \sin C\cos \theta - \cos C\sin \theta }}$
Simplifying this, we’ll get:
$
   \Rightarrow \dfrac{{\cos A + \cos B}}{{\cos A - \cos B}} = \dfrac{{2\sin C\cos \theta }}{{ - 2\cos C\sin \theta }} \\
   \Rightarrow \dfrac{{\cos A + \cos B}}{{\cos A - \cos B}} = - \dfrac{{\tan C}}{{\tan \theta }} \\
 $
Here are another two trigonometric formulas we are going to use next:
\[
   \Rightarrow \cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right) \\
   \Rightarrow \cos C - \cos D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{D - C}}{2}} \right) \\
 \]
Using these formulas on the left hand side of the equation, we’ll get:
$ \Rightarrow \dfrac{{2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)}}{{2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{B - A}}{2}} \right)}} = - \dfrac{{\tan C}}{{\tan \theta }}$
We know that $\sin \left( { - \theta } \right) = - \sin \theta $, we can substitute $\sin \left( {\dfrac{{B - A}}{2}} \right) = - \sin \left( {\dfrac{{A - B}}{2}} \right)$. Doing this and simplifying expressions, we’ll get:
$
   \Rightarrow \dfrac{{\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)}}{{ - \sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)}} = - \dfrac{{\tan C}}{{\tan \theta }} \\
   \Rightarrow - \dfrac{1}{{\tan \left( {\dfrac{{A + B}}{2}} \right)\tan \left( {\dfrac{{A - B}}{2}} \right)}} = - \dfrac{{\tan C}}{{\tan \theta }} \\
 $
Cross multiplying the equation, we’ll get:
$ \Rightarrow \tan \theta = \tan \left( {\dfrac{{A + B}}{2}} \right)\tan \left( {\dfrac{{A - B}}{2}} \right)\tan C$

Option (B) is the correct answer.

Note: When two ratios are equal i.e. $\dfrac{a}{b} = \dfrac{c}{d}$ then we can use componendo and dividend as shown below:
$ \Rightarrow \dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}{\text{ }}.....{\text{(1)}}$
We can use only componendo:
$ \Rightarrow \dfrac{{a + b}}{b} = \dfrac{{c + d}}{d}{\text{ }}.....{\text{(1)}}$
Or we can use only dividendo:
$ \Rightarrow \dfrac{{a - b}}{b} = \dfrac{{c - d}}{d}{\text{ }}.....{\text{(3)}}$
Dividing equation (2) and (3), we’ll get componendo and dividend. This is what we have used in the above problem.