Q.In $\triangle ABC$, prove that $\tan\left(\dfrac{C-A}{2}\right) = \left(\dfrac{c-a}{c+a}\right)\cot\dfrac{B}{2}$.

Write $a,c$ via the sine rule ($a=k\sin A$, $c=k\sin C$), apply sum-to-product formulas to $\sin C \pm \sin A$, and use $A+B+C=\pi$.

By the sine rule, $\dfrac{a}{\sin A}=\dfrac{c}{\sin C}=k$ (say), so $a=k\sin A$, $c=k\sin C$.

$\frac{c-a}{c+a} = \frac{k\sin C - k\sin A}{k\sin C+k\sin A} = \frac{\sin C-\sin A}{\sin C+\sin A}$

Using sum-to-product identities:

$\sin C - \sin A = 2\cos\left(\frac{C+A}{2}\right)\sin\left(\frac{C-A}{2}\right)$

$\sin C + \sin A = 2\sin\left(\frac{C+A}{2}\right)\cos\left(\frac{C-A}{2}\right)$

So:

$\frac{c-a}{c+a} = \frac{\cos\left(\frac{C+A}{2}\right)\sin\left(\frac{C-A}{2}\right)}{\sin\left(\frac{C+A}{2}\right)\cos\left(\frac{C-A}{2}\right)} = \tan\left(\frac{C-A}{2}\right)\cot\left(\frac{C+A}{2}\right)$

Since $A+B+C=\pi$, we have $\dfrac{C+A}{2} = \dfrac{\pi-B}{2} = \dfrac{\pi}{2}-\dfrac{B}{2}$, so

$\cot\left(\frac{C+A}{2}\right) = \cot\left(\frac{\pi}{2}-\frac{B}{2}\right) = \tan\left(\frac{B}{2}\right)$

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