Question

$(b^2-c^2)\cot A+(c^2-a^2)\cot B+(a^2-b^2)\cot C=0$.

Answer 1

$(b^2-c^2)\cot A=(b^2-c^2)\frac{\cos A}{\sin A}$
$=(b^2-c^2).\frac{b^2+c^2-a^2}{2ba.ka}$
$\frac{1}{2k.abc}[(b^4-c^4)-a^2(b^2-c^2\left)\right]$
$\therefore L.H.S=\dfrac { 1 }{ 2k.abc } \left[ \left( { b }^{ 4 }-c^{ 4 } \right) +\left( c^{ 4 }-a^{ 4 } \right) +\left( a^{ 4 }-
b^{ 4 } \right) -\left\{ { a }^{ 2 }\left( { b }^{ 2 }-{ c }^{ 2 } \right) +{ b }^{ 2 }\left( c^{ 2 }-a^{ 2 } \right) +c^{ 2 }\left( {
a }^{ 2 }-{ b }^{ 2 } \right) \right\} \right] =0$.