# prove that

• December 3rd 2009, 01:04 AM
bigwave
prove that
prove that $b^2(cotA + cotB) = c^2(cotA + cotC)$

$b,c$ being sides of a triangle
$A,B,C$ being corresponding interior angles
• December 3rd 2009, 05:44 AM
alexmahone
Quote:

Originally Posted by bigwave
prove that $b^2(cotA + cotB) = c^2(cotA + cotC)$

$b,c$ being sides of a triangle
$A,B,C$ being corresponding interior angles

$\frac{cot A+cot C}{cot A+cot B}=\frac{\frac{cos A}{sin A}+\frac{cos C}{sin C}}{\frac{cos A}{sin A}+\frac{cos B}{sin B}}$

= $\frac{sin (A+C)}{sin A sin C}*\frac{sin A sin B}{sin(A+B)}$

= $\frac{sin (\pi-B)}{sin C}*\frac{sin B}{sin(\pi-C)}$

= $\frac{sin B}{sin C}*\frac{sin B}{sin C}$

= $\frac{sin^2B}{sin^2C}$

= $\frac{b^2}{c^2}$

Thus, $b^2(cotA + cotB) = c^2(cotA + cotC)$