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Question

(b2c2)cotA+(c2a2)cotB+(a2b2)cotC=0.

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Solution

(b2c2)cotA=(b2c2)cosAsinA
=(b2c2).b2+c2a22ba.ka
12k.abc[(b4c4)a2(b2c2)]
$\therefore L.H.S=\dfrac { 1 }{ 2k.abc } \left[ \left( { b }^{ 4 }-c^{ 4 } \right) +\left( c^{ 4 }-a^{ 4 } \right) +\left( a^{ 4 }-
b^{ 4 } \right) -\left\{ { a }^{ 2 }\left( { b }^{ 2 }-{ c }^{ 2 } \right) +{ b }^{ 2 }\left( c^{ 2 }-a^{ 2 } \right) +c^{ 2 }\left( {
a }^{ 2 }-{ b }^{ 2 } \right) \right\} \right] =0$.

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