If secA + cosA = 5/3, find secA - cosA
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Answered by
4
seca +cosa=5/3
squaring it we get
sec^2a+cos^2a= 25/9-2
let us substract 2 from both sides
sec^a+cos^2a-2=25/9-4
seca-cosa =√25/9-4
squaring it we get
sec^2a+cos^2a= 25/9-2
let us substract 2 from both sides
sec^a+cos^2a-2=25/9-4
seca-cosa =√25/9-4
Answered by
3
Given
Sec A + cos A=5/3--(1)
(a-b)^2 = (a+b)^2-4ab
(secA-cosA)^2=(secA +cosA)^2-4secAcosA
(secA-cosA)^2= (5/3)^2-1 here we used secAcosA=1
=25/9 -1
=(25-9)/9
=16/9
=(4/3)^2
Therefore
SecA - cosA = 4/3 or -4/3
Sec A + cos A=5/3--(1)
(a-b)^2 = (a+b)^2-4ab
(secA-cosA)^2=(secA +cosA)^2-4secAcosA
(secA-cosA)^2= (5/3)^2-1 here we used secAcosA=1
=25/9 -1
=(25-9)/9
=16/9
=(4/3)^2
Therefore
SecA - cosA = 4/3 or -4/3
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