SOLUTION: How do I verify these two identities? Here are the two identities.
1. (sin(x)sec^2(x) + sin(x) - 2tan(x)) / (cos(x) - 1)^2 = sec(x)tan(x)
2. (sin^3(x)-cos^3(x)) / (sin(x) - c
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-> SOLUTION: How do I verify these two identities? Here are the two identities.
1. (sin(x)sec^2(x) + sin(x) - 2tan(x)) / (cos(x) - 1)^2 = sec(x)tan(x)
2. (sin^3(x)-cos^3(x)) / (sin(x) - c
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Question 535932: How do I verify these two identities? Here are the two identities.
1. (sin(x)sec^2(x) + sin(x) - 2tan(x)) / (cos(x) - 1)^2 = sec(x)tan(x)
2. (sin^3(x)-cos^3(x)) / (sin(x) - cos(x)) = 1 + sin(x)cos(x) Answer by Aswathy(23) (Show Source):
You can put this solution on YOUR website! (1)
LHS =[sinx sec^2x + sinx - 2tanx] / (cosx - 1)^2 (tanx=sinx/cosx=secx/cosecx)
=[sinx(sec^2x + 1- 2secx)]/[(1/secx-1]^2
=[sinx(sec^2x + 1- 2secx)] / [(1-secx)/secx]^2
=[sinx(sec^2x + 1- 2secx)] / [(1-2secx+ sec^2x)/sec^2x]
=[sinx(sec^2x + 1- 2secx)]
=sinx*sec^2x [cancelling out (sec^2x + 1- 2secx)]
=sinx*1/cosx*secx (tanx=sinx/cosx=secx/cosecx)
=tanx*secx=secx*tanx=RHS
(2)
LHS =[sin^3(x)-cos^3(x)] /[sin(x) - cos(x)]
=[(sinx-cosx)(sin^2x+sinxcosx+cos^2x)] / (sinx-cosx)
=sin^2+sinxcosx+cos^2x {cancelling out (sinx-cosx)]
=1+sinxcosx (using identity sin^2+cos^2x=1)
=RHS