sinA +cosA = root3 than find tanA + cotA = 1
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Answered by
216
SinA + cosA = √3
Squaring on both sides we get,
(SinA + cosA)² = (√3)²
Sin²A + cos²A +2sinAcosA = 3
1 + 2sinAcosA = 3
2sinAcosA = 3-1
SinAcosA = 2/2
sinAcosA = 1......................(!)
tanA+cotA = 1
sinA/cosA + cosA/sinA = 1
sin²A + cos²A /sinAcosA = 1
1/sinAcosA = 1
sinAcosA = 1.....................(!!)
! = !!
thus tanA + cotA = 1
Squaring on both sides we get,
(SinA + cosA)² = (√3)²
Sin²A + cos²A +2sinAcosA = 3
1 + 2sinAcosA = 3
2sinAcosA = 3-1
SinAcosA = 2/2
sinAcosA = 1......................(!)
tanA+cotA = 1
sinA/cosA + cosA/sinA = 1
sin²A + cos²A /sinAcosA = 1
1/sinAcosA = 1
sinAcosA = 1.....................(!!)
! = !!
thus tanA + cotA = 1
what?
Continuataiom......sinAcosA = 1(Since sin²A + cos²A= 1 SinAcosA= sin²A + cos²A Divide both sides by sinAcosA sinAcosA/sinAcosA= sin²A/sinAcosA + cos²A/sinAcosA Then, 1= sinA/cosA+ cosA/sinA Then, 1= TanA+ Cot A Hence Proved..........................
This is Correct
Apt
then y did u post the question?
We got the answer just now
hmm
thnks
Answered by
38
given that sinA+cosA=√3
tanA+cotA=sin^2A+cos^2A/sinAcosA=1/sinAcosA
(sinA+cosA)^2=1+2sinAcosA.........(1)
given that sinA+cosA=√3
(1)⇒3-1/2=sinAcosA=1
hence proved
tanA+cotA=sin^2A+cos^2A/sinAcosA=1/sinAcosA
(sinA+cosA)^2=1+2sinAcosA.........(1)
given that sinA+cosA=√3
(1)⇒3-1/2=sinAcosA=1
hence proved
Similar questions
Squaring on both sides we get,
(SinA + cosA)² = (√3)²
Sin²A + cos²A +2sinAcosA = 3
1 + 2sinAcosA = 3
2sinAcosA = 3-1
SinAcosA = 2/2
sinAcosA = 1......................(!)