If 0

Question

If 0 < a < b < c, then cot-1((ab+1/a-b))+cot-1((bc+1/b-c))+cot-1((ca+1/c-a))= (A) 0 (B) π (C) 2π (D) None of these

Tardigrade Admin · Accepted Answer

∵ a-b < 0, so cot-1 (ab+1/a-b)=cot-1 b-cot-1 a+π b-c < 0, so cot-1 (bc+1/b-c)=cot-1 c-cot-1 b+π c-b < 0, so cot-1 (ca+1/c-a)=cot-1 a-cot-1 c Adding we get cot-1((ab+1/a-b))+cot-1((bc+1/b-c))+cot-1((ca+1/c-a))=2π

Q. If $0 < a < b < c$ , then
$co t^{- 1} (\frac{ab + 1}{a - b}) + co t^{- 1} (\frac{b c + 1}{b - c}) + co t^{- 1} (\frac{c a + 1}{c - a}) =$

$0$

$π$

$2 π$

None of these

Q. If 0<a<b<c, then cot−1(a−bab+1​)+cot−1(b−cbc+1​)+cot−1(c−aca+1​)=

0

π

2π

None of these

∵a−b<0, so cot−1a−bab+1​=cot−1b−cot−1a+π b−c<0, so cot−1b−cbc+1​=cot−1c−cot−1b+π c−b<0, so cot−1c−aca+1​=cot−1a−cot−1c Adding we get cot−1(a−bab+1​)+cot−1(b−cbc+1​)+cot−1(c−aca+1​)=2π

Q. If $0 < a < b < c$ , then
$co t^{- 1} (\frac{ab + 1}{a - b}) + co t^{- 1} (\frac{b c + 1}{b - c}) + co t^{- 1} (\frac{c a + 1}{c - a}) =$

$0$

$π$

$2 π$