Question

in any triangleABC P.T

(b2-c2)cotA + (c2-a2)cotB + (a2-b2)cotC = 0

Answer 1

This question is based on properties of triangle .
we know, $\bold{\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}=P[{\text{constant}}]}$
so, sinA = aP , sinB = bP , sinC = cP
also we know,
$\bold{cosA=\frac{b^2+c^2-a^2}{2bc}}\\\\\bold{cosB=\frac{c^2+a^2-b^2}{2ca}}\\\\\bold{cosC=\frac{a^2+b^2-c^2}{2ab}}$
so, $\bold{cotA=\frac{b^2+c^2-a^2}{2abcP}}\\\\\bold{cotB=\frac{c^2+a^2-b^2}{2abcP}}\\\\\bold{cotC=\frac{a^2+b^2-c^2}{2abcP}}$

Now, LHS = (b² - c²)cotA + (c² - a²)cotB + (a² - b²)cotC
Putting the values of cotA , cotB and cotC
= 1/2abcP[ (b² - c²)(b² + c² - a²) + (c² - a²)(c² + a² - b²) + (a² - b²)(a² + b² - c²)]
= 1/2abcP [ b⁴ - c⁴ -a²b² + a²c² + c⁴ - a⁴ -b²c² + b²a² + a⁴ - b⁴ - c²a² + c²b² ]
= 1/2abcP × 0 = RHS

Hence , proved

Answer 2

what he have given is the answer for your question and their p is 2R

where R is the circumradius