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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.

Answers(3)

Answer

Parthasaradhi M

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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