Question

If $\cos A=m\operatorname{Cos}B$, then
A . $\cot \dfrac{A+B}{2}=\dfrac{m+1}{m-1}\tan \dfrac{B-A}{2}$
B . $\cot \dfrac{A+B}{2}=\dfrac{m+1}{m-1}\cot \dfrac{B-A}{2}$
C . $\cot \dfrac{A+B}{2}=\dfrac{m+1}{m-1}\tan \dfrac{A-B}{2}$
D . None of these

Answer 1

Hint: In this question, we are given that $\cos A=m\operatorname{Cos}B$. First we find the value of m from the given equation. Then we apply componendo and dividendo on the equation which is sometimes in short we called C And D rule and by using the formulas of $\cos A+\cos B$ and $\cos A-\cos B$and putting the formulas in the equation and simplifying it, we are able to get the desirable answer and choose the correct option.

Formula Used:
In this question, we use the identity which are described below :-
$\cos A+\cos B=2\cos \dfrac{A+B}{2}\cos \dfrac{B-A}{2}$
And $\cos A-\cos B=2\sin \dfrac{A+B}{2}\sin \dfrac{B-A}{2}$

Complete step- by- step Solution:
Given $\cos A=m\operatorname{Cos}B$
Then $\dfrac{\cos A}{\cos B}=m$
Now we apply componendo and dividendo in the above equation,
Componendo and dividendo is a theorem on proportions which allows for an easy way to perform calculations and helps in reducing the number of expansions needed.
We get
$\dfrac{\cos A+\cos B}{\cos A-\cos B}=\dfrac{m+1}{m-1}$……………………………… (1)
Now we use the trigonometric formulas according to the above equation, which is
We know $\cos A+\cos B=2\cos \dfrac{A+B}{2}\cos \dfrac{B-A}{2}$
And $\cos A-\cos B=2\sin \dfrac{A+B}{2}\sin \dfrac{B-A}{2}$
Put the above formulas in equation (1), we get
$\dfrac{2\cos \dfrac{A+B}{2}\cos \dfrac{B-A}{2}}{2\sin \dfrac{A+B}{2}\sin \dfrac{B-A}{2}}=\dfrac{m+1}{m-1}$
By simplifying the above equation, we get
$\dfrac{\cot \dfrac{A+B}{2}}{\tan \dfrac{B-A}{2}}=\dfrac{m+1}{m-1}$
Now we multiply $\tan \dfrac{B-A}{2}$ on both sides, we get
$\cot \dfrac{A+B}{2}=\dfrac{m+1}{m-1}\tan \dfrac{B-A}{2}$

Thus, Option ( A ) is the correct answer.

Note: Componendo and dividendo is a mathematical rule that states that the product of two or more binomials can be divided by the sum of the individual terms within each binomial. This allows for simplified algebraic equations and solutions. The dividendo portion of the equation belongs to the first binomial, while the componendo belongs to all subsequent binomials.