Question

$\large cotA + cotB + cotC = \frac { a^{2} + b^{2} + c^{2}}{4∆}$​

Answer 1

we have to prove , cotA+cotB + cotC = a²+b²+c²/4∆

We know cosine rule

, cosA = b²+c²-a²/2bc

CosB = a²+c²-b²/2ac

cosC = a²+b²-c²/2ab

CotA = cosA/sinA = b²+c²-a²/2bcsinA

CotB = cosB/sinB =a²+c²-b²/2casinB

cotC = cosC/sinC = a²+b²-c²/2absinC

We also know that,

∆ = 1/2bcsinA = 1/2acsinB = 1/2absinC

since , cotA +cotB+cotC

=> b²+c²-a²/2bcsinA + a²+c²-b²/2acsinB + a²+b²-c²/2absinC

Multiply by 2 and divide by 2 we get

=> b²+c²-a²/4∆ + a²+c²-b²/4∆ + a²+b²-c²/4∆

=> a²+b²+c²/4∆ Proved

_________________________

Answer 2

your answer mate

see this attachment