# Show that (b^2-c^2)*cotA+(c^2-a^2)*cotB+(a^2-b^2)*cotC=0?

Jan 16, 2018

By sine law we know

$\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C = 2 R$

Now
1st part

$\left({b}^{2} - {c}^{2}\right) \cot A$

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

=4R^2(1/2(1-cos2B)-1/2(1-cos2C)cotA

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

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

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

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

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

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

Similarly

2nd part $= \left({c}^{2} - {a}^{2}\right) \cot B$

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

3rd part $= \left({a}^{2} - {b}^{2}\right) \cot C$

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

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