in any triangleABC P T (b2-c2)cotA + (c2-a2)cotB + (a2-b2)cotC = 0 - Maths - Trigonometric Functions

Question

in any triangleABC P T
(b2-c2)cotA +
(c2-a2)cotB +
(a2-b2)cotC = 0

Ankush Jain · Accepted Answer

Let $\frac{a}{\sin â¡ â€‰ A} = \frac{b}{\sin â¡ â€‰ B} = \frac{c}{\sin â¡ â€‰ C} = k$ Then a = ksin A, b = ksin B, c = ksin C $L H S = (b^{2} - c^{2}) \cot A + (c^{2} - a^{2}) \cot B + (a^{2} - b^{2}) \cot C = (b^{2} - c^{2}) \frac{\cos A}{\sin A} + (c^{2} - a^{2}) \frac{\cos B}{\sin B} + (a^{2} - b^{2}) \frac{\cos C}{\sin C} = \frac{(b^{2} - c^{2})}{k a} [\frac{b^{2} + c^{2} - a^{2}}{2 b c}] + \frac{(c^{2} - a^{2})}{k b} [\frac{a^{2} + c^{2} - b^{2}}{2 a c}] + \frac{(a^{2} - b^{2})}{k c} [\frac{a^{2} + b^{2} - c^{2}}{2 a b}]$ $= \frac{1}{2 k a b c} [(b^{2} - c^{2}) (b^{2} + c^{2} - a^{2}) + (c^{2} - a^{2}) (a^{2} + c^{2} - b^{2}) + (a^{2} - b^{2}) (a^{2} + b^{2} - c^{2})] = \frac{1}{2 k a b c} [(b^{2} - c^{2}) (b^{2} + c^{2}) - a^{2} (b^{2} - c^{2}) + (c^{2} - a^{2}) (c^{2} + a^{2}) - b^{2} (c^{2} - a^{2}) + (a^{2} - b^{2}) (a^{2} + b^{2}) - c^{2} (a^{2} - b^{2})] = \frac{1}{2 k a b c} [(b^{2} - c^{2}) (b^{2} + c^{2}) + (c^{2} - a^{2}) (c^{2} + a^{2}) + (a^{2} - b^{2}) (a^{2} + b^{2}) - a^{2} (b^{2} - c^{2}) - b^{2} (c^{2} - a^{2}) - c^{2} (a^{2} - b^{2})] = \frac{1}{2 k a b c} [(b^{4} - c^{4}) + (c^{4} - a^{4}) + (a^{4} - b^{4}) - (a^{2} b^{2} - a^{2} c^{2}) - (b^{2} c^{2} - b^{2} a^{2}) - (c^{2} a^{2} - c^{2} b^{2})] = \frac{1}{2 k a b c} \times 0 = 0 = RHS$

in any triangleABC P.T (b2-c2)cotA + (c2-a2)cotB + (a2-b2)cotC = 0

in any triangleABC P.T

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