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Homework Help: Need help proving some trig identities

  1. Aug 9, 2014 #1
    Proving identities is a pain! Thanks in advance, guys!

    1. The problem statement, all variables and given/known data

    1. 1 + sec^(2)xsin^(2)x = sec^(2)x

    2. sinx/1-cosx + sinx/1+cosx = 2cscx

    2. Relevant equations

    3. The attempt at a solution

    For the first problem, this is the best I got:

    1 + sec^(2)x(1-cos(2)x)

    For the second problem, I added the fractions together and got:

    (sinx + sinxcosx + sinx - sinxcosx) / ((1-cosx) (1-cosx)) =

    (2sinx) / (1-cosx) (1+cosx) =

    (2sinx) / (1-cos^(2)x) =

    (2sinx) / (sin^(2)x)
  2. jcsd
  3. Aug 9, 2014 #2


    Staff: Mentor

    Welcome to PF

    For the first one, go back to basics and replace sec with its cos definition ie sec(x)=1/cos(x) and it should be apparent.

    For the second one, try combining the fractions on the left and also replace csc(x) with its sin(x) definition.
  4. Aug 10, 2014 #3
    Hmmm I kinda see what you're saying. I started all over with the first one and this is what I got, can anyone check if I did it correctly? It proves to be correct, at least to me.

    1 + (1/cos^(2)x)(sin^(2))x) = I then multiplied the two fractions, giving me:

    1 + (sin^(2)x)/(cos^(2)x) = now I combine the fractions, giving me:

    (cos^(2)x + sin^(2)x)/(cos^(2)x) = now I apply a Pythagorean identity to the numerator:

    (1)/(cos^(2)x) = sec^(2)x

    I am still stumped on the second one. Are you saying combine fractions from the initial equation or the one I worked on and got? Thanks
  5. Aug 10, 2014 #4


    User Avatar
    Homework Helper

    Combine the LHS of number 2. Do you see why that is a good idea?
  6. Aug 10, 2014 #5
    I think I got it now. For number 2, I separated the two fractions from the left side of the initial equation and got this:

    (sinx)/(1) - (sinx)/(cosx) + (sinx)/(1) +(sinx)/(cosx) = then I combined similar terms and got this:

    (sinx)/(1) + (sinx)/(1) = now I flipped the fractions and got this:

    (1)/(sinx) + (1)/(sinx) = (2)/(sinx) = 2cscx

    Can anyone check if I did it correctly? Thanks
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