# 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

Algebra ->  Algebra  -> Trigonometry-basics -> 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      Log On

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 Algebra: Trigonometry Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Trigonometry-basics 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