If $b-a=mc$ then prove that $\cot (\dfrac {B-A}{2})=\dfrac {1+m.\cos (B)}{m.sin (B)}$ If $b-a=mc$ then prove that $\cot (\dfrac {B-A}{2})=\dfrac {1+m.\cos (B)}{m.sin (B)}$
My Attempt:
We know,
$$\tan (\dfrac {B-A}{2} )=\dfrac {b-a}{b+a}.\cot (\dfrac {C}{2})$$.
Then
$$b-a=mc$$
$$(b+a).\tan (\dfrac {B-A}{2}).\cot (\dfrac {C}{2})=mc$$

Since (b-a) = mc
m = (b-a)/c
L.H.S = $\frac{1+\frac{\left(b-a\right)}{c}CosB}{\frac{(b-a)}{c}SinB}$
= $\frac{c+\left(b-a\right)CosB}{\left(b-a\right)SinB}$
$=\frac{SinC+\left(SinB-SinA\right)CosB}{ \begin{array}{c}
\left(SinB-SinA\right)SinB \\ 
 \end{array}
}
$
$ =\frac{{\mathrm{sin} \left(A+B\right)\ }-SinA.CosB+SinB.CosB}{ \begin{array}{c}
{Sin}^2B-SinA.SinB \\ 
 \end{array}
}$
$ =\frac{CosASinB+SinBCosB}{ \begin{array}{c}
{Sin}^2B-SinA.SinB \\ 
 \end{array}
}$
$ =\frac{CosA+CosB}{SinB-SinA}$
=$cot\left(\frac{B-A}{2}\right)$ = R.H.S
