Question

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

Answer 1

Answer:

2 is power or what.......

Answer 2

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

Step-by-step explanation:

By sine rule, we have

$\dfrac{a}{\sin A} =\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

⇒ $\dfrac{\sin A}{a} =\dfrac{\sin B}{b} =\dfrac{\sin C}{c} =k$

⇒ $\sin A=ka, \sin B=kb$ and $\sin C=kc$

∴ L.H.S. $=(b^2-c^2)\cot A + (c^2-a^2)\cot B + (a^2-b^2)\cot C$

$=(b^2-c^2)\dfrac{\cos A}{\sin A} + (c^2-a^2)\dfrac{\cos B}{\sin B} + (a^2-b^2)\dfrac{\cos C}{\sin C}$

$=\dfrac{b^2-c^2}{ka}.\dfrac{b^2+c^2-a^2}{2bc}+\dfrac{c^2-a^2}{kb}.\dfrac{c^2+a^2-b^2}{ca} + \dfrac{a^2-b^2}{kc}.\dfrac{a^2+b^2-c^2}{2ab}$

$=\dfrac{1}{2kabc}[(b^2-c^2)(b^2+c^2-a^2)+(c^2-a^2)(c^2+a^2-b^2)+(a^2-b^2)(a^2+b^2-c^2)]$

$=\dfrac{1}{2kabc}[0]$

= 0 = R.H.S., proved,