If abc$$0thencot$$−$$1ab+1a$$−$$b+$$$$cot$$−$$1bc+1b$$−$$c+$$$$cot$$−$$1ca+1c$$−$$a=$$

Question

If a>b>c>$$0thencot$$−$$1ab+1a$$−$$b+$$$$cot$$−$$1bc+1b$$−$$c+$$$$cot$$−$$1ca+1c$$−$$a=$$

Infinity Learn · Accepted Answer

Since a>b>c>$$0$$, we have $$a-b$$>$$0$$, $$b-c$$>$$0$$ and $$c-a$$<$$0$$  ∴$$cot$$−$$1ab+1a$$−$$b+$$$$cot$$−$$1bc+1b$$−$$c+$$$$cot$$−$$1ca+1c$$−$$a=$$$$cot$$−$$1ab+1a$$−$$b+$$$$cot$$−$$1bc+1b$$−$$c+$$$$cot$$−$$1-ca+1a-c$$                                                                                          $$=$$$$cot$$$$-1a-$$$$cot$$$$-1b+$$$$cot$$$$-1b-$$$$cot$$$$-1c+$$π$$-$$$$cot$$$$-1a-$$$$cot$$$$-1c$$                                                                                                                                       ∵$$cot$$$$-1-x=$$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>0thencot−1ab+1a−b+cot−1bc+1b−c+cot−1ca+1c−a=

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