Results 1 to 7 of 7

Math Help - Establish the trig identity

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    2

    Establish the trig identity

    \frac{\sec\theta - 1}{\tan^3\theta} = \frac{\cot\theta}{\sec\theta + 1}

    I need help verifying that these two expressions are equal (getting one side of the equation to look like the other without using properties of equality, just trig identities and simplifying one side).

    Thanks.
    Last edited by rebelspartan; October 8th 2009 at 07:07 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Hello rebelspartan

    Welcome to Math Help Forum!
    Quote Originally Posted by rebelspartan View Post
    \frac{\sec\theta - 1}{\tan^3\theta} = \frac{\cot\theta}{\sec\theta + 1}

    I need help verifying that these two expressions are equal (getting one side of the equation to look like the other without using properties of equality, just trig identities and simplifying one side).

    Thanks.
    Three things you need here:

    • \sec^2\theta = 1+\tan^2\theta (1)


    • \cot\theta = \frac{1}{\tan\theta} (2)


    • a^2-b^2 = (a+b)(a-b) (3)

    So:

    1. Express the denominator (of the LHS) as \tan\theta\times\tan^2\theta, and then use (1) to express \tan^2\theta in terms of \sec^2\theta
    2. Use (3) to factorise the denominator
    3. Cancel (simplify) the whole fraction
    4. Use (2) on the remaining term in \tan\theta

    ... and you're there.

    Can you do that?

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    2
    Great, I got it! Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member pacman's Avatar
    Joined
    Jul 2009
    Posts
    448


    rebelspartan, this is less rigorous but, what if we cross multiply? let t = theta

    --------------------------------------------------------------------------------------------------------------

    (sec t - i)(sec t + 1) = cot t tan^3 t = (1/tan t)(tan t)(tan^2 t), that red thing cancels out.

    the LHS is the difference of 2 squares, (sec t - 1)(sec t + 1) = sec^2 t - 1

    LHS = RHS, indeed sec^2 t - 1 = tan^2 t,

    remember this PYTHAGOREAN IDENTITY: tan^2 t + 1 = sec^2 t
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Hello everyone
    Quote Originally Posted by pacman View Post


    rebelspartan, this is less rigorous but, what if we cross multiply? let t = theta

    --------------------------------------------------------------------------------------------------------------

    (sec t - i)(sec t + 1) = cot t tan^3 t = (1/tan t)(tan t)(tan^2 t), that red thing cancels out.

    the LHS is the difference of 2 squares, (sec t - 1)(sec t + 1) = sec^2 t - 1

    LHS = RHS, indeed sec^2 t - 1 = tan^2 t,

    remember this PYTHAGOREAN IDENTITY: tan^2 t + 1 = sec^2 t
    Very bad style! Don't start off by assuming the thing you want to prove, and end up proving something that is known to be true. This is completely the wrong way round!

    Having said that, if you can do this, then you should be able to 'reverse-engineer' your proof so that it all reads in the right direction and is therefore mathematically sound.

    Grandad
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member pacman's Avatar
    Joined
    Jul 2009
    Posts
    448
    i agree with you grandad, my above computation is really mathematically unsound, because i assumed it already as an identity, but i experimented with it by hindsight . . . . and it works.

    ,

    formal Proof:

    Consider the RHS, that is (cot t)/(sec t + 1) only

    multiply and divide by (sec t - 1)

    (cot t)/(sec t + 1) = [(cot t)/(sec t + 1)][(sec t - 1)/(sec t - 1)],

    use the identity: 1 + tan^2 t = sec^2 t and simplify,

    (cot t)/(sec t + 1) = (cot t)(sec t - 1)/(sec^2 t - 1)

    (cot t)/(sec t + 1) = (cot t)(sec t - 1)/(tan^2 t)

    (cot t)/(sec t + 1) = (cot t)(sec t - 1)/(tan^2 t)

    (cot t)/(sec t + 1) = (1/tan t)(sec t - 1)/(tan^2 t)

    (cot t)/(sec t + 1) = (sec t - 1)/[(tan t)(tan^2 t)]

    (cot t)/(sec t + 1) = (sec t - 1)/(tan^3 t) = LHS.

    -
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Hello pacman

    That's much better!

    Grandad
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with De Moivre formula to establish identity
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: December 24th 2011, 10:48 AM
  2. establish the following:
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: March 26th 2010, 11:12 AM
  3. establish the identity
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: November 1st 2009, 04:50 PM
  4. establish the identity
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: April 24th 2009, 03:56 AM
  5. Establish the Identity
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: August 25th 2008, 08:54 PM

Search Tags


/mathhelpforum @mathhelpforum