If abc$$0$$,then $$cot$$−$$1ab+1a$$−$$b+$$$$cot$$−$$1bc+1b$$−$$c+$$$$cot$$−$$1ca$$−$$1c$$−a

Question

If  a>b>c>$$0$$,then $$cot$$−$$1ab+1a$$−$$b+$$$$cot$$−$$1bc+1b$$−$$c+$$$$cot$$−$$1ca$$−$$1c$$−a

Infinity Learn · Accepted Answer

Given a>b>c>$$0$$.  Now $$cot$$$$-1ab+1a-b+$$$$cot$$$$-1bc+1b-c+$$$$cot$$$$-1ca+1c-a=$$$$cot$$$$-1ab+1a-b+$$$$cot$$$$-1bc+1b-c+$$$$cot$$$$-1-ac+1a-c$$                                                                                      $$=$$$$cot$$$$-1a-$$$$cot$$$$-1b+$$$$cot$$$$-1b-$$$$cot$$$$-1c+$$π$$-$$$$cot$$$$-1a-$$$$cot$$$$-1c$$                                                                                             ∵if x>y, then $$cot$$$$-1xy+1x-y=$$$$cot$$$$-1x-$$$$cot$$$$-1y$$,  also $$cot$$$$-1-x=$$π$$-$$$$cot$$$$-1x$$                                                                                      $$=$$π

If $a > b > c > 0,$ then $\cot^{- 1} (\frac{a b + 1}{a - b}) + \cot^{- 1} (\frac{b c + 1}{b - c}) + \cot^{- 1} (\frac{c a - 1}{c - a})$

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If a>b>c>0,then cot−1ab+1a−b+cot−1bc+1b−c+cot−1ca−1c−a

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If $a > b > c > 0,$ then $\cot^{- 1} (\frac{a b + 1}{a - b}) + \cot^{- 1} (\frac{b c + 1}{b - c}) + \cot^{- 1} (\frac{c a - 1}{c - a})$