Question

If $\cos A = m\cos B$ and $\cot \dfrac{{A + B}}{2} = \lambda \tan \dfrac{{B - A}}{2}$, then $\lambda $ is
(A) $\dfrac{m}{{m - 1}}$
(B) $\dfrac{{m + 1}}{m}$
(C) $\dfrac{{m + 1}}{{m - 1}}$
(D) None of these

Answer 1

Hint:
Taking the first equation can bring ‘m’ on one side and cosine terms on others. Now use the method of componendo and dividendo to form a new expression, i.e. $\dfrac{a}{b} = \dfrac{c}{d} \Rightarrow \dfrac{{a + b}}{{a - b}} = \dfrac{{c + d}}{{c - d}}$ . Now on RHS, use the cosine and sine addition angle formulas. Rearrange the expression to change the ratio of sine and cosine into tangent and cotangent. Now compare the obtained equation with $\cot \dfrac{{A + B}}{2} = \lambda \tan \dfrac{{B - A}}{2}$ to find the required value.

Complete step by step solution:
Here in the given problem, we have two trigonometric equations $\cos A = m\cos B$ and $\cot \dfrac{{A + B}}{2} = \lambda \tan \dfrac{{B - A}}{2}$. And using these two equations, we need to find the value for unknown $'\lambda '$ . According to the options we need to express the value of $'\lambda '$ in terms of \['m'\].
Let’s take the first equation and try to transform it into some useful form.
$ \Rightarrow \cos A = m\cos B \Rightarrow m = \dfrac{{\cos A}}{{\cos B}}$ …………(i)
Now let’s use the method of componendo and dividendo in the above equation. If $a$ ,$b$ ,$c$ and $d$ are numbers such that $b$ and $d$ are non-zero and $\dfrac{a}{b} = \dfrac{c}{d}$ , then the following holds:
${\text{Componendo: }}\dfrac{{a + b}}{b} = \dfrac{{c + d}}{d}$
${\text{Dividendo: }}\dfrac{{a - b}}{b} = \dfrac{{c - d}}{d}$
So, let’s use the methods of componendo and dividendo in the relation (i), we get:
$ \Rightarrow m = \dfrac{{\cos A}}{{\cos B}} \Rightarrow \dfrac{{m + 1}}{{m - 1}} = \dfrac{{\cos A + \cos B}}{{\cos A - \cos B}}$
Now, for RHS, we can use the addition formula for sine and cosine, i.e. $\cos M + \cos N = 2\cos \left( {\dfrac{{M + N}}{2}} \right)\cos \left( {\dfrac{{M - N}}{2}} \right)$ and $\cos M - \cos N = 2\sin \left( {\dfrac{{M + N}}{2}} \right)\sin \left( {\dfrac{{M - N}}{2}} \right)$
 $ \Rightarrow m = \dfrac{{\cos A}}{{\cos B}} \Rightarrow \dfrac{{m + 1}}{{m - 1}} = \dfrac{{\cos A + \cos B}}{{\cos A - \cos B}} = \dfrac{{2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{B - A}}{2}} \right)}}{{2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{B - A}}{2}} \right)}}$
Now, $2$ can be cancel from both numerator and numerator
$ \Rightarrow \dfrac{{m + 1}}{{m - 1}} = \dfrac{{\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{B - A}}{2}} \right)}}{{\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{B - A}}{2}} \right)}}$
This above expression can be rearranged further as:
$ \Rightarrow \dfrac{{m + 1}}{{m - 1}} = \dfrac{{\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{B - A}}{2}} \right)}}{{\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{B - A}}{2}} \right)}} = \dfrac{{\dfrac{{\cos \left( {\dfrac{{A + B}}{2}} \right)}}{{\sin \left( {\dfrac{{A + B}}{2}} \right)}}}}{{\dfrac{{\sin \left( {\dfrac{{B - A}}{2}} \right)}}{{\cos \left( {\dfrac{{B - A}}{2}} \right)}}}}$
Since we know that the ratio of sine by cosine is tangent and the reciprocal of the tangent is cotangent, i.e. $\tan M = \dfrac{{\sin M}}{{\cos M}}$ and $\cot M = \dfrac{{\cos M}}{{\sin M}}$ . Let’s use this in the above relationship:
$ \Rightarrow \dfrac{{m + 1}}{{m - 1}} = \dfrac{{\dfrac{{\cos \left( {\dfrac{{A + B}}{2}} \right)}}{{\sin \left( {\dfrac{{A + B}}{2}} \right)}}}}{{\dfrac{{\sin \left( {\dfrac{{B - A}}{2}} \right)}}{{\cos \left( {\dfrac{{B - A}}{2}} \right)}}}} = \dfrac{{\cot \left( {\dfrac{{A + B}}{2}} \right)}}{{\tan \left( {\dfrac{{B - A}}{2}} \right)}}$
Therefore, we get: \[ \Rightarrow \tan \left( {\dfrac{{B - A}}{2}} \right) = \dfrac{{m + 1}}{{m - 1}}\cot \left( {\dfrac{{A + B}}{2}} \right)\] …………….(ii)
So, now we can compare the given equation, which is $\cot \dfrac{{A + B}}{2} = \lambda \tan \dfrac{{B - A}}{2}$ , by the above-formed relation (ii). By doing this we can get the value of $\lambda $ in terms of $'m'$ as:
$ \Rightarrow \lambda = \dfrac{{m + 1}}{{m - 1}}$

Hence, the option (C) is the correct answer.

Note:
In questions like this, the use of proper trigonometric identity is very crucial. An alternative approach is to take the equation $\cot \dfrac{{A + B}}{2} = \lambda \tan \dfrac{{B - A}}{2}$ and try to transform it into the form of cosine angles of ‘A’ and ‘B’. Then you can use the equation one to find the required value.