किसी `triangleABC` में सिद्ध करे कि
`((a + b+ c)^(2))/(a^(2) + b^(2) + c^(2)) = (cot""(A)/(2)+ cot""(B)/(2) +cot""(C)/(2))/(cotA + cotB + cotC)`

Explanation

दायाँ पक्ष का अंश ` = cot ""(A)/(2) + cot""(B)/(2) + cot""(C)/(2)`
` = (s(s-a))/(Delta) + (s(s-b))/(Delta) + (s(s-b))/(Delta)`
` = (s)/(Delta) [(s-a) + (s-b) + (s-c)]`
` = (s)/(Delta) [3s -(a+b+c)] = (s)/(Delta) (3s-2s) = (s^(2))/(Delta)`
दायाँ पक्ष का हर ` = cotA + cotB + cotC`
` = (cosA)/(sinA) + (cosB)/(sinB) + (cosC)/(sinC)`
` = (b^(2) + c^(2) - a^(2))/(2bca) *2R + (c^(2) + a^(2) - b^(2))/(2cab) *2R + (a^(2) + b^(2) - c^(2))/(2abc) *2R`
` [because (a)/(sinA) = (b)/(sinB) = (c)/(sinC) = 2R]`
`= (R)/(abc) [b^(2) + c^(2) -a^(2)) + (c^(2) + a^(2) - b^(2)) + (a^(2) + b^(2) - c^(2))]`
` = (R)/(abc) [a^(2) + b^(2) + c^(2)] = (a^(2) + b^(2) + c^(2))/(4Delta) [because Delta = (abc)/(4R) therefore abc = 4DeltaR]`

अब दायाँ पक्ष ` = (cot""(A)/(2) + cot""(B)/(2) + cot""(C)/(2))/(cotA + cotB =cotC) = (s^(2))/(Delta) xx (4Delta)/(a^(2) + b^(2) + c^(2))`
` = ((2s)^(2))/(a^(2) + b^(2) + c^(2)) = ((a + b + c)^(2))/(a^(2) + b^(2) + c^(2)) =` बायाँ पक्ष