किसी `Delta ABC` में सिध्द कीजिए -
`cot.(A)/(2) + cot.(B)/(2)+cot.(C )/(2) =(a+b+c)/(b+c+a)cot.(A)/(2)`

Explanation

बायाँ पक्ष `= cot. (A)/(2)+cot.(B)/(2)+cot.(C )/(2)`
`=sqrt([(s(s-a))/((s-b)(s-c))]]+sqrt([(s(s-b))/((s-c)(s-a))]] +sqrt([(s(s-c))/((s-a)(s-b))])`
`(sqrt(s)[(s-a)+(s-b)+(s-c)])/(sqrt([(s-a)(s-b)(s-c)]]]= (sqrt(3)[3s -(a+b+c)])/(sqrt[[(s-a)(s-b)(s-c)]]]`
`=(sqrt(s))/(sqrt([(s-b)(s-c)]]]=((3s-2s))/(sqrt((s-a)))=sqrt([(s-(s-a))/((s-b)(s-c))]] xx (s)/(s-a)`
`=cot.(A)/(2)xx (2s)/(2s-2a)=cot.(A)/(2)xx (a+b+c)/((a+b+c)-2a)`
` = (a+b+c)/(b+c-a) cot. (A)/(2)`= दायाँ पक्ष।