1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    jedishrfu

    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

    verty

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Need help proving some trig identities
Loading...