Results 1 to 4 of 4

Thread: Trig Indentity Help (Proof)

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    3

    Trig Indentity Help (Proof)

    $\displaystyle [cot X/(1-tan X)] + [tan X/(1-cot X] = 1+ tan X + cot X$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Thanks
    1
    Awards
    1
    I tried this with no luck... tried multiplying out by the conjugates and then LCD's, but nothing...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,656
    Thanks
    14
    Quote Originally Posted by viciouspoultry View Post
    $\displaystyle [cot X/(1-tan X)] + [tan X/(1-cot X] = 1+ tan X + cot X$
    $\displaystyle \frac{{\cot x}}
    {{1 - \tan x}} + \frac{{\tan x}}
    {{1 - \cot x}} = \frac{{\dfrac{{\cos x}}
    {{\sin x}}}}
    {{\dfrac{{\cos x - \sin x}}
    {{\cos x}}}} - \frac{{\dfrac{{\sin x}}
    {{\cos x}}}}
    {{\dfrac{{\cos x - \sin x}}
    {{\sin x}}}}.$

    Now this is equal to $\displaystyle \frac{{\cos ^2 x}}
    {{\sin x(\cos x - \sin x)}} - \frac{{\sin ^2 x}}
    {{\cos x(\cos x - \sin x)}}.$ Factorise:

    $\displaystyle \frac{1}
    {{\cos x - \sin x}}\left\{ {\frac{{\cos ^2 x}}
    {{\sin x}} - \frac{{\sin ^2 x}}
    {{\cos x}}} \right\}.$ After some simple calculations we have

    $\displaystyle \frac{1}
    {{\cos x - \sin x}}\left\{ {\frac{{(\cos x - \sin x)\left( {\cos ^2 x + \sin x\cos x + \sin ^2 x} \right)}}
    {{\sin x\cos x}}} \right\},$ and we happily get

    $\displaystyle 1 + \tan x + \cot x,$ as required $\displaystyle \blacksquare$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, viciouspoultry!

    Thought I'd try this one head-on . . .


    $\displaystyle \frac{\cot x}{1-\tan x}+\frac{\tan x}{1-\cot x} \;= \;1+ \tan x + \cot x$

    On the left, multiply the second fraction by $\displaystyle \frac{\tan x}{\tan x}\!:\;\;\;\frac{\cot x}{1 -\tan x} \:+\:{\color{blue}\frac{\tan x}{\tan x}}\cdot\frac{\tan x}{1-\cot x}$

    . . $\displaystyle =\;\;\frac{\cot x}{1-\tan x} + \frac{\tan^2\!x}{\tan x - 1}\;\;=\;\;\frac{\cot x}{1-\tan x} - \frac{\tan^2\!x}{1-\tan x} \;\;=\;\; \frac{\cot x-\tan^2\!x}{1-\tan x} $


    Multiply by $\displaystyle \frac{\tan x}{\tan x}\!:\;\;\;{\color{blue}\frac{\tan x}{\tan x}}\cdot\frac{\cot x - \tan^2\!x}{1-\tan x} \;=\; \frac{1 - \tan^3\!x}{\tan x(1-\tan x)}$


    Factor and reduce: .$\displaystyle \frac{(1-\tan x)(1+ \tan x+ \tan^2\!x)}{\tan x(1-\tan x)} \;=\;\frac{1+\tan x+\tan^2\!x}{\tan x}$


    Then we have: . $\displaystyle \frac{1}{\tan x} + \frac{\tan x}{\tan x} + \frac{\tan^2\!x}{\tan x}\;\;=\;\; \cot x + 1 +\tan x$

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Trig Indentity Limits
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Jun 2nd 2010, 10:20 PM
  2. prov trig indentity
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: Jan 12th 2010, 02:21 PM
  3. trig indentity proof help,
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Dec 15th 2009, 12:58 PM
  4. please check my working, trig indentity proof
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: Dec 14th 2009, 08:22 AM
  5. Trig indentity
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: May 2nd 2006, 01:30 PM

Search Tags


/mathhelpforum @mathhelpforum