Explanation

(c)
We have, $\cos{A = m\cos B}$
$\Rightarrow \frac{\cos A}{\cos B} = \frac{m}{1}$
$\Rightarrow \frac{\cos{A + \cos B}}{\cos{A - \cos B}} = \frac{m + 1}{m - 1}$
$\Rightarrow \ \frac{2\cos{\frac{A + B}{2}\cos\frac{B - A}{2}}}{2\sin{\frac{A + B}{2}\sin\frac{B - A}{2}}} = \frac{m + 1}{m - 1}$
$\Rightarrow \cot{\frac{A + B}{2} = \left( \frac{m + 1}{m - 1} \right)\tan\frac{B - A}{2}}$
But${\ cot}{\frac{A + B}{2} = \lambda}\tan\frac{B - A}{2}$
$\therefore\ \lambda = \frac{m + 1}{m - 1}$