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cot(A-B).Cot(B-C)+Cot(C-A).Cot(B-C)+Cot(C-A).Cot(A-B)=1

cot(A-B).Cot(B-C)+Cot(C-A).Cot(B-C)+Cot(C-A).Cot(A-B)=1

Grade:10

1 Answers

Arnab
26 Points
3 years ago
let, x= A-B
      y= B-C
      z= C-A
\therefore x+y+z= A-B+B-C+C-A
\therefore x+y+z= 0
\therefore x+y= -z
\therefore cot(x+y)= cot(-z)
\therefore (cot x • cot y - 1)/(cot y + cot x)= -cot z
\therefore cot x• cot y -1 = -cot z(cot y + cot x)
\therefore cot x• cot y -1 = -cot z• cot y - cot z• cot x
\therefore cot x• cot y + cot z• cot y + cot z• cot x= 1
\therefore cot(A-B).cot(B-C)+cot(C-A).cot(B-C)+cot(C-A)+cot(A-B)=1
                             hence proved.

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