If $cos A = m cos B$ , then

cot (\frac{A + B}{2}) = \frac{m + 1}{m - 1} tan (\frac{B - A}{2})

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tan (\frac{A + B}{2}) = \frac{m + 1}{m - 1} cot (\frac{B - A}{2})

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cot (\frac{A + B}{2}) = \frac{m + 1}{m - 1} tan (\frac{A - B}{2})

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tan (\frac{A - B}{2}) = \frac{m + 1}{m - 1} cot (\frac{B - A}{2})

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Solution

The correct option is B $tan (\frac{A + B}{2}) = \frac{m + 1}{m - 1} cot (\frac{B - A}{2})$
$\begin{matrix} cos A = m cos B \frac{cos A}{cos B} = m \to (i) O n a d d i n g a n d s u b s t r a c t i n g b y o n e i n e q u a t i o n (i) b o t h s i d e s \Rightarrow \frac{cos A + cos B}{cos B} = m + 1 \Rightarrow \frac{2 cos (\frac{A + B}{2}) \cdot cos (\frac{A - B}{2})}{cos B} = m + 1 \to (i i) \Rightarrow \frac{- 2 sin (\frac{A + B}{2}) \cdot sin (\frac{A - B}{2})}{(cos B)} = (m - 1) \to (i i i) d i v i d e e q u a t i o n (i i i) b y (i i) tan (\frac{A + B}{2} \cdot) tan (\frac{A - B}{2}) = \frac{(1 + m)}{(1 - m)} \end{matrix}$
$tan \frac{A + B}{2} = \frac{m + 1}{m - 1} cot \frac{B - A}{2}$

Hence, this is the answer.

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