Results 1 to 3 of 3

Math Help - identity proving with two terms in denominator

  1. #1
    Newbie
    Joined
    Mar 2013
    From
    Maryland
    Posts
    3

    identity proving with two terms in denominator

    I have a trig identity proving question:

    (1 - sin(x))/(1 + sin(x)) = tan(x) + 1/cos(x)

    by multiplying 1 - sin(x) on left side doesn't seems work.

    Anyone can help. Thanks in advanced.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,739
    Thanks
    645

    Re: identity proving with two terms in denominator

    Hello, ellicott!

    Are there typos?
    The statement is not true.


    \text{Prove: }\:\frac{1 - \sin x}{1 + \sin x} \:=\: \tan x + \frac{1}{\cos x}

    \text{Let }x = \tfrac{\pi}{6}

    \text{The left side is: }\:\frac{1-\sin\frac{\pi}{6}}{1+\sin\frac{\pi}{6}} \:=\:\frac{1-\frac{1}{2}}{1 + \frac{1}{2}} \;=\;\frac{\frac{1}{2}}{\frac{3}{2}} \;=\;\tfrac{1}{3}

    \text{The right side is: }\:\tan\tfrac{\pi}{6} + \sec\tfrac{\pi}{6} \;=\;\tfrac{1}{\sqrt{3}} + \tfrac{2}{\sqrt{3}} \;=\;\tfrac{3}{\sqrt{3}} \;=\;\sqrt{3}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    I found this identity . . .


    \frac{1-\sin x}{1+\sin x} \;=\; \frac{1 -\sin x}{1+\sin x}\cdot {\color{blue}\frac{1-\sin x}{1-\sin x}} \;=\;\frac{(1-\sin x)^2}{1 - \sin^2\!x} \;=\;\frac{(1-\sin x)^2}{\cos^2\!x}

    . . . . . . =\;\left(\frac{1-\sin x}{\cos x}\right)^2 \;=\;\left(\frac{1}{\cos x} - \frac{\sin x}{\cos x}\right)^2 \;=\;(\sec x - \tan x)^2


    Therefore: . \frac{1-\sin x}{1+\sin x} \;=\;(\sec x - \tan x)^2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2013
    From
    Maryland
    Posts
    3

    Re: identity proving with two terms in denominator

    Thanks Soroban for correction of my typo. In general, what is the method to test true identity? Do you have to use a particular angle such as pi/6?

    appreciate your helps,
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving a log identity
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 16th 2009, 07:16 PM
  2. Help proving this identity?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 30th 2009, 02:40 PM
  3. Proving an identity that's proving to be complex
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 21st 2009, 01:30 PM
  4. Proving identity #1
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: April 9th 2009, 02:33 PM
  5. Proving an identity
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: November 14th 2008, 10:55 AM

Search Tags


/mathhelpforum @mathhelpforum