36. If K = (seca + tan a)(sec 3 + tan 3)(sec y +tan y)= (seca – tan a)(sec 3 — tan 8) (secay – tan).
then K =
a) 0

December 26, 2021

By Hailey

1 thought on “36. If K = (seca + tan a)(sec 3 + tan 3)(sec y +tan y)= (seca – tan a)(sec 3 — tan 8) (secay – tan).<br /> then K =<br /> a) 0<br”

1. Answer:

   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

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