Let a triangle ABC of sides a,b,c, having area Delta
Area =∆ ‌
1
2
bc‌sin‌A=∆....(i) and ‌‌‌
1
2
ac‌sin‌B=∆ ....(ii) and ‌‌‌
1
2
ab‌sin‌c=∆....(iii) Using cosine rule a2=b2+c2−2bc‌cos‌A and ‌‌b2=a2+c2−2ac‌cos‌B and ‌‌c2=a2+b2−2ab‌cos‌C On adding, we get
a2+b2+c2=2a2+2b2+2c2−2ab‌cos‌C−2ac‌cos‌B−2bc‌cos‌A or a2+b2+c2=2(ab‌cos‌C+ac‌cos‌B+bc‌cos‌A)...(iv)
Now, from Eq. (i) bc=‌
2∆
sin‌A
Eq. (ii), ac=‌
2∆
sin‌B
Eq. (iii), ‌‌ab=‌
2∆
sin‌C
Putting these values in Eq. (iv), we get
a2+b2+c2=2(‌
2∆
sin‌C
‌cos‌C+‌
2∆
sin‌B
‌cos‌B+‌
2∆
sin‌A
‌cos‌A) a2+b2+c2=4∆(cot‌C+cot‌B+cot‌A) or cot‌C+cot‌B+cot‌A=‌