Results 1 to 4 of 4

Thread: prove (tan@+sec@-1)/(tan@-sec@+1)=(1+sin@)/(cos@) by method different from book's

  1. #1

    Question prove (tan@+sec@-1)/(tan@-sec@+1)=(1+sin@)/(cos@) by method different from book's

    1. I have been trying to prove this trigonometric identity:

    . . . . .[tex]\dfrac{\tan(\theta)\, +\, \sec(\theta)\, -\, 1}{\tan(\theta)\, -\, \sec(\theta)\, +\, 1}\, =\, \dfrac{1\, +\, \sin(\theta)}{\cos(\theta)}[/tex]

    My textbook proves it using two different methods. See them here.

    Now, I made an attempt to solve this question in a different way. I have somehow reached this step:

    . . . . .[tex]\mbox{L.H.S. }\, =\, \dfrac{\tan(\theta)\, +\, \sec(\theta)\, -\, 1}{\tan(\theta)\, -\, \sec(\theta)\, +\, 1}\, =\, \dfrac{\sqrt{\sin(\theta)\, +\, 1\,}}{\sqrt{1\, -\, \cos(\theta)\,}}\, \cdot\, \dfrac{\sqrt{1\, +\, \sin(\theta)\,}\, -\, \sqrt{1\, -\, \sin(\theta)\,}}{\sqrt{1\, +\, \cos(\theta)\,}\, -\, \sqrt{1\, -\, \cos(\theta)\,}}[/tex]

    But, I am stuck in this position. Please help me to solve it further. Please tell, how can I prove the rest?

    . . . . .[tex]\dfrac{\sqrt{\sin(\theta)\, +\, 1\,}}{\sqrt{1\, -\, \cos(\theta)\,}}\, \cdot\, \dfrac{\sqrt{1\, +\, \sin(\theta)\,}\, -\, \sqrt{1\, -\, \sin(\theta)\,}}{\sqrt{1\, +\, \cos(\theta)\,}\, -\, \sqrt{1\, -\, \cos(\theta)\,}}\, =\, \dfrac{1\, +\, \sin(\theta)}{\cos(\theta)}[/tex]

    I also want to know one more thing,

    2. By how many different methods can a single trigonometric identity be proved? (one, two, or infinitely many)

    I will be thankful for help!
    Last edited by stapel; 03-03-2018 at 08:16 PM. Reason: Typing out the text in the graphics; creating useful subject line.

  2. #2
    Junior Member
    Join Date
    Jan 2018
    Location
    Boston
    Posts
    131
    If A/B = C/D then A*D = B*C. Prove the second equality instead. Seems like this approach was not used.

    Regarding your other question, there can be many proofs. It's difficult to say how many. How would you formally classify them? Is proof #50 a new approach or just a slight tweak of proof #35? Is this question from the book?

  3. #3
    Elite Member
    Join Date
    Apr 2005
    Location
    USA
    Posts
    9,340
    There are many proofs for the Pythagorean Theorem. Here are 118. https://www.cut-the-knot.org/pythagoras/
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  4. #4
    Elite Member
    Join Date
    Jun 2007
    Posts
    17,454
    Quote Originally Posted by navneet9431 View Post
    1. I have been trying to prove this trigonometric identity:

    . . . . .[tex]\dfrac{\tan(\theta)\, +\, \sec(\theta)\, -\, 1}{\tan(\theta)\, -\, \sec(\theta)\, +\, 1}\, =\, \dfrac{1\, +\, \sin(\theta)}{\cos(\theta)}[/tex]

    My textbook proves it using two different methods. See them here.

    Now, I made an attempt to solve this question in a different way. I have somehow reached this step:

    . . . . .[tex]\mbox{L.H.S. }\, =\, \dfrac{\tan(\theta)\, +\, \sec(\theta)\, -\, 1}{\tan(\theta)\, -\, \sec(\theta)\, +\, 1}\, =\, \dfrac{\sqrt{\sin(\theta)\, +\, 1\,}}{\sqrt{1\, -\, \cos(\theta)\,}}\, \cdot\, \dfrac{\sqrt{1\, +\, \sin(\theta)\,}\, -\, \sqrt{1\, -\, \sin(\theta)\,}}{\sqrt{1\, +\, \cos(\theta)\,}\, -\, \sqrt{1\, -\, \cos(\theta)\,}}[/tex]

    But, I am stuck in this position. Please help me to solve it further. Please tell, how can I prove the rest?

    . . . . .[tex]\dfrac{\sqrt{\sin(\theta)\, +\, 1\,}}{\sqrt{1\, -\, \cos(\theta)\,}}\, \cdot\, \dfrac{\sqrt{1\, +\, \sin(\theta)\,}\, -\, \sqrt{1\, -\, \sin(\theta)\,}}{\sqrt{1\, +\, \cos(\theta)\,}\, -\, \sqrt{1\, -\, \cos(\theta)\,}}\, =\, \dfrac{1\, +\, \sin(\theta)}{\cos(\theta)}[/tex]

    I also want to know one more thing,

    2. By how many different methods can a single trigonometric identity be proved? (one, two, or infinitely many)

    I will be thankful for help!
    I am Khan-fused with all those √ signs in your work.

    If I were to do this problem, I would start as follows:

    [tex]\displaystyle{\dfrac{tan\theta + sec\theta - 1}{tan\theta - sec\theta + 1}}[/tex]

    =[tex]\displaystyle{\dfrac{sin\theta - cos\theta + 1}{sin\theta + cos\theta - 1}}[/tex]

    =[tex]\displaystyle{\dfrac{sin\theta - cos\theta + 1}{sin\theta + cos\theta - 1} * \dfrac{sin\theta + (1 - cos\theta)}{sin\theta + (1 - cos\theta)}}[/tex]

    =[tex]\displaystyle{\dfrac{(sin\theta - cos\theta + 1)^2}{sin^2\theta - (1- cos\theta)^2}}[/tex]

    and continue.....
    Last edited by stapel; 03-03-2018 at 08:17 PM. Reason: Copying typed-out graphical content into reply.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

Tags for this Thread

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •