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Nitin Rajda Rajda

Sep 21, 2013

# If (secA + tanA)(secB + tanB)(secC + tan C) = (secA - tanA)(secB - tanB)(secC - tanC)

If (secA + tanA)(secB + tanB)(secC + tan C) = (secA - tanA)(secB - tanB)(secC - tanC), prove that each of the sides is equal to 1 or -1.

Member since Apr 1, 2017

Solution: Given: (secA + tanA)(secB + tanB)(secC + tan C) = (secA - tanA)(secB - tanB)(secC - tanC) Multiplying both sides with by LHS i.e  "(secA + tanA)(secB + tanB)(secC + tan C) " ⇒ (secA + tanA)(secB + tanB)(secC + tan C)(secA + tanA)(secB + tanB)(secC + tan C) = (secA - tanA)(secB - tanB)(secC - tanC)(secA + tanA)(secB + tanB)(secC + tan C)  ⇒ (secA + tanA)2(secB + tanB)2(secC + tan C)2 = (sec2A - tan2A)(sec2B - tan2B)(sec2C - tan2C)            [∵  (a+b)(a-b) = a2 - b2 ] ⇒ [(secA + tanA)(secB + tanB)(secC + tan C)]2  = (1)(1)(1)  = 1                                       [∵  sec2A - tan2A = 1 ] ∴ (secA + tanA)(secB + tanB)(secC + tan C) = ± 1 Similarly  Multiplying both sides with by RHS i.e " (secA - tanA)(secB - tanB)(secC - tanC) " we will get ⇒ (sec2A - tan2A)(sec2B - tan2B)(sec2C - tan2C) = (secA -  tanA)2(secB - tanB)2(secC -  tan C)2 ⇒ (1)(1)(1) = [(secA - tanA)(secB - tanB)(secC - tan C)]2          [∵  sec2A - tan2A = 1 ]  ⇒ [(secA - tanA)(secB - tanB)(secC - tan C)]2  = 1 ∴ (secA – tanA)(secB – tanB)(secC – tan C) = ± 1

Agam Gupta

Consider,(secA + tanA)(secB + tanB)(secC + tan C) = (secA - tanA)(secB - tanB)(secC - tanC) Mulitply both sides with "(secA + tanA)(secB + tanB)(secC + tan C)", we get (secA + tanA)2(secB + tanB)2(secC + tan C)2�= (sec2A - tan2A)(sec2B - tan2B)(sec2C - tan2C) ����������������������������������������������������������������������� = (1)(1)(1) = 1 ⇒ [(secA + tanA)(secB + tanB)(secC + tan C)]2�= 1 ∴ (secA + tanA)(secB + tanB)(secC + tan C) = ± 1 Similarly, we get (secA – tanA)(secB – tanB)(secC – tan C) = ± 1

Kishore Kumar

Consider,(secA + tanA)(secB + tanB)(secC + tan C) = (secA - tanA)(secB - tanB)(secC - tanC) Mulitply both sides with "(secA + tanA)(secB + tanB)(secC + tan C)", we get (secA + tanA)2(secB + tanB)2(secC + tan C)2 = (sec2A - tan2A)(sec2B - tan2B)(sec2C - tan2C)                                                                         = (1)(1)(1) = 1 ⇒ [(secA + tanA)(secB + tanB)(secC + tan C)]2 = 1 ∴ (secA + tanA)(secB + tanB)(secC + tan C) = ± 1 Similarly, we get (secA – tanA)(secB – tanB)(secC – tan C) = ± 1
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