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!

Verify the identity

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Verify the Identity:

    cos(x)-[cos(x)/1-tan(x)] = [(sin(x)cos(x)]/[sin(x)-cos(x)]


    b]2. Relevant equations[/b]
    reciprocal Identities, quotient Identities, Pythagorean Identities


    3. The attempt at a solution
    cos(x)-[cos(x)/1-tan(x)] = [(sin(x)cos(x)]/[sin(x)-cos(x)]

    into

    [cos(x)(1-tan(x))-cos(x)]/[1-tan(x)]

    to

    [-cos(x)tan(x)]/[1-tan(x)]


    and this is where i get stuck cant turn it to [(sin(x)cos(x)]/[sin(x)-cos(x)]
    i hope i wrote the problem right
     
  2. jcsd
  3. Feb 21, 2010 #2



    Try to multiply [tex]\frac{cos(x)}{cos(x)}[/tex] to the very initial equation on the left-hand side , thus combine the terms into a fraction.
     
  4. Feb 21, 2010 #3
    when i do that wont i get [-cos^2(x)sin(x)]/[cos(x)-sin(x)]

    then what...sorry I'm not that good as these kinda stuff
     
  5. Feb 21, 2010 #4
    [tex]cos(x)-\frac{cos^{2}(x)}{cos(x)-sin(x)}[/tex]
    Now how do you make the denominator of the fraction as sin(x)-cos(x) ?
    [hint: multiply -1]
     
  6. Feb 21, 2010 #5
    [tex]cos(x)-\frac{cos^{2}(x)}{cos(x)-sin(x)}[/tex]

    after long starring and thinking a light bulb just lit in my head lol

    so...

    [tex]\frac{\cos^2(x)-\sin(x)\cos(x)-\cos^2(x)}{cos(x)-sin(x)}[/tex]

    to

    [tex]\frac{-\sin(x)\cos(x)}{cos(x)-sin(x)}[/tex] then all multiplied by -1 equals...

    [tex]\frac{\sin(x)\cos(x)}{sin(x)-cos(x)}[/tex]

    am i right? i hope im right...
     
  7. Feb 21, 2010 #6

    Yes . You're right!
     
  8. Feb 21, 2010 #7
    thank you so much for helping me :biggrin:
     
  9. Feb 21, 2010 #8

    Mark44

    Staff: Mentor

    On a point of terminology, the left-hand side is an expression that is part of an equation, but it's not an equation. It is incorrect to refer to an equation on either side of an equation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Verify the identity
  1. Verifying the Identity (Replies: 1)

  2. Verifying identity (Replies: 2)

  3. Verifying an Identity (Replies: 9)

  4. Verifying an Identity (Replies: 2)

  5. Verify the identity (Replies: 1)

Loading...