SecA + cosecA = 2 then find sec^4 + cosec^4
Answers
Given : SecA + cosecA = 2
To find : sec⁴A + Cosec⁴A
Solution:
SecA + cosecA = 2
=> 1/cosA + 1/sinA = 2
=> sinA + CosA = 2SinACosA
=> Sin²A + Cos²A + 2SinACosA = 4Sin²ACos²A
=> 1 + 2SinACosA = 4Sin²ACos²A
=> 4x² - 2x - 1 = 0
SinACosA = x
=> 4x² - 2x - 1 = 0
=> x = (2 ± √20)/8
=> x = (1 ± √5)/2
=> SinACosA = (1 ± √5)/2
=> SecAcosecA = 2/(1 ± √5)
SecA + cosecA = 2
Squaring both sides
=> sec²A + Cosec²A + 2SecAcosecA = 4
=> sec²A + Cosec²A = 4 - 2SecAcosecA
Squaring both sides
=> sec⁴A + Cosec⁴A + 2 sec²A Cosec²A = 16 + 4Sec²Acosec²A - 16secAcosecA
=> sec⁴A + Cosec⁴A = 16 + 2Sec²Acosec²A - 16secAcosecA
16 + 2Sec²Acosec²A - 16secAcosecA
SecAcosecA = 2/(1 ± √5)
putting this value we can get sec⁴A + Cosec⁴A
16 + 2 ( 2/(1 ± √5))² - 16(2/(1 ± √5))
Learn more:
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