Question

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

Answer 1

In $\Delta ABC$ by \sine rule, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

$\therefore a=k\sin A,b=k\sin B,c=k\sin C$

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

$=\frac{\sin C-\sin A}{\sin C+\sin A}$

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

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

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

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