Results 1 to 3 of 3

Math Help - trig proof

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    1

    trig proof

    prove that

    (1-sinA)/(1-secA)-(1+sinA)/(1+secA)=2cotA(cosA-cosecA)

    if anyone can show me how to do this it would be greatly appreciated
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602
    Quote Originally Posted by thehaanz View Post
    prove that

    (1-sinA)/(1-secA)-(1+sinA)/(1+secA)=2cotA(cosA-cosecA)

    if anyone can show me how to do this it would be greatly appreciated
    \frac{1 - \sin{A}}{1 - \sec{A}} - \frac{1 + \sin{A}}{1 + \sec{A}} = \frac{1 - \sin{A}}{1 - \frac{1}{\cos{A}}} - \frac{1 + \sin{A}}{1 + \frac{1}{\cos{A}}}

     = \frac{1 - \sin{A}}{\frac{\cos{A} - 1}{\cos{A}}} - \frac{1 + \sin{A}}{\frac{\cos{A} + 1}{\cos{A}}}

     = \frac{\cos{A}(1 - \sin{A})}{\cos{A} - 1} - \frac{\cos{A}(1 + \sin{A})}{\cos{A} + 1}

     = \frac{\cos{A}(\cos{A} + 1)(1 - \sin{A})}{(\cos{A} - 1)(\cos{A} + 1)} - \frac{\cos{A}(\cos{A} - 1)(1 + \sin{A})}{(\cos{A} - 1)(\cos{A} + 1)}

     = \frac{\cos{A}(\cos{A} + 1)(1 - \sin{A}) - \cos{A}(\cos{A} - 1)(1 + \sin{A})}{\cos^2{A} - 1}

     = \frac{\cos{A}(\cos{A} - 1)(1 + \sin{A}) - \cos{A}(\cos{A} + 1)(1 - \sin{A})}{\sin^2{A}}

     = \frac{\cos{A}[(\cos{A} - 1)(1 + \sin{A}) - (\cos{A} + 1)(1 - \sin{A})]}{\sin^2{A}}

     = \frac{\cos{A}(\cos{A} + \cos{A}\sin{A} - 1 - \sin{A} - \cos{A} + \cos{A}\sin{A} - 1 + \sin{A})}{\sin^2{A}}

     = \frac{\cos{A}(2\cos{A}\sin{A} - 2)}{\sin^2{A}}

     = \frac{2\cos{A}(\cos{A}\sin{A} - 1)}{\sin^2{A}}

     = 2\cot{A}\left(\frac{\cos{A}\sin{A} - 1}{\sin{A}}\right)

     = 2\cot{A}\left(\frac{\cos{A}\sin{A}}{\sin{A}} - \frac{1}{\sin{A}}\right)

     = 2\cot{A}(\cos{A} - \csc{A}).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,904
    Thanks
    765
    Hello, thehaanz!

    Prove that: . \frac{1-\sin A}{1-\sec A} -\frac{1+\sin A}{1+\sec A} \;=\;2\cot A\,(\cos A- \csc A)
    Start with the left side . . .


    \frac{1-\sin A}{1-\sec A} - \frac{1+\sin A}{1 + \sec A} \;\;=\;\;\frac{(1-\sin A)(1+\sec A) - (1+\sin A)(1-\sec A)}{(1-\sec A)(1 + \sec A)}

    . . . . . . . . . . . . . . =\;\;\frac{1 + \sec A - \sin A - \sin A\sec A - 1 + \sec A - \sin A + \sin A\sec A}{1-\sec^2A}

    . . . . . . . . . . . . . . =\;\;\frac{2\sec A - 2\sin A}{-(\sec^2\!A - 1)} \;\;=\;\;\frac{-2(\sin A - \sec A)}{-\tan^2\!A} \;\;=\;\;\frac{2(\sin A - \sec A)}{\tan^2\!A}

    . . . . . . . . . . . . . . =\;\;\frac{2\left(\sin A - \dfrac{1}{\cos A}\right)}{\dfrac{\sin^2\!A}{\cos^2\!A}} \;\;=\;\;2\cdot\frac{\cos^2\!A}{\sin^2\!A}\left(\s  in A - \frac{1}{\cos A}\right)

    . . . . . . . . . . . . . . =\;\;2\cdot\frac{\cos A}{\sin A}\cdot\underbrace{\frac{\cos A}{\sin A}\left(\sin A - \frac{1}{\cos A}\right)}

    . . . . . . . . . . . . . . =\quad 2\cdot\frac{\cos A}{\sin A}\cdot\left(\cos A - \frac{1}{\sin A}\right)

    . . . . . . . . . . . . . . =\qquad 2\cdot\cot A(\cos A - \csc A)

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig Proof
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: October 27th 2010, 06:55 PM
  2. Trig proof for sec^2
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: July 12th 2010, 12:40 AM
  3. trig proof
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: March 5th 2009, 11:57 PM
  4. Help with a Trig Proof
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 11th 2008, 12:16 PM
  5. trig proof
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: September 14th 2008, 07:30 PM

Search Tags


/mathhelpforum @mathhelpforum