Question

Prove that $(b-c)\cot \frac{1}{2}A+(c-a)\cot \frac{1}{2}B+(a-b)\cot \frac{1}{2}C=0$.

Answer 1

L.H.S

$(b-c)\cot \frac{A}{2}+(c-a)\cot \frac{B}{2}+(a-b)\cot \frac{C}{2}$ ......... (1)

We know that $sine$ rule,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Now, $1$st Part

$(b-c)\cot \frac{A}{2}$

$=(4R^2{\sin }^{2}B-4R^2{\sin }^{2}C)\cot \frac{A}{2}$

$=4R^2(\frac{1-\cos 2B}{2}-\frac{1-\cos 2C}{2})\cot \frac{A}{2}$

$=4R^2\times \frac{1}{2}(\cos 2C-\cos 2B)\cot \frac{A}{2}$

$=2R^2\times (\cos 2C-\cos 2B)\cot \frac{A}{2}$

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

$=4R^2\times \left(\sin \right(\pi -A\left)\sin \right(B-C\left)\right)\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}$

$=4R^2\times \left(\sin A\sin \right(B-C\left)\right)\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}$

$=4R^2\sin (B-C)\cos \frac{A}{2}$

$(b-c)\cot \frac{A}{2}=4R^2[\sin B\cos C-\cos B\sin C]\cos \frac{A}{2}$ ......... (2)

Similarly, $2$nd Part

$(c-a)\cot \frac{B}{2}=4R^2[\sin C\cos A-\cos C\sin A]\cos \frac{B}{2}$ ......... (3)

Similarly, $3$rd Part

$(a-b)\cot \frac{C}{2}=4R^2[\sin A\cos B-\cos A\sin B]\cos \frac{C}{2}$ ......... (4)

On adding equation (2), (3) and (4), we get

$(b-c)\cot \frac{A}{2}+(c-a)\cot \frac{B}{2}+(a-b)\cot \frac{C}{2}=0$

Hence, proved.