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Math Help - Trig Indentity Help (Proof)

  1. #1
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    Trig Indentity Help (Proof)

    [cot X/(1-tan X)]  +  [tan X/(1-cot X]   =   1+ tan X + cot X
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  2. #2
    GAMMA Mathematics
    colby2152's Avatar
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    I tried this with no luck... tried multiplying out by the conjugates and then LCD's, but nothing...
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  3. #3
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    Quote Originally Posted by viciouspoultry View Post
    [cot X/(1-tan X)]  +  [tan X/(1-cot X]   =   1+ tan X + cot X
    \frac{{\cot x}}<br />
{{1 - \tan x}} + \frac{{\tan x}}<br />
{{1 - \cot x}} = \frac{{\dfrac{{\cos x}}<br />
{{\sin x}}}}<br />
{{\dfrac{{\cos x - \sin x}}<br />
{{\cos x}}}} - \frac{{\dfrac{{\sin x}}<br />
{{\cos x}}}}<br />
{{\dfrac{{\cos x - \sin x}}<br />
{{\sin x}}}}.

    Now this is equal to \frac{{\cos ^2 x}}<br />
{{\sin x(\cos x - \sin x)}} - \frac{{\sin ^2 x}}<br />
{{\cos x(\cos x - \sin x)}}. Factorise:

    \frac{1}<br />
{{\cos x - \sin x}}\left\{ {\frac{{\cos ^2 x}}<br />
{{\sin x}} - \frac{{\sin ^2 x}}<br />
{{\cos x}}} \right\}. After some simple calculations we have

    \frac{1}<br />
{{\cos x - \sin x}}\left\{ {\frac{{(\cos x - \sin x)\left( {\cos ^2 x + \sin x\cos x + \sin ^2 x} \right)}}<br />
{{\sin x\cos x}}} \right\}, and we happily get

    1 + \tan x + \cot x, as required \blacksquare
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  4. #4
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    Hello, viciouspoultry!

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


    \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 \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}

    . . =\;\;\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 \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: . \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: . \frac{1}{\tan x} + \frac{\tan x}{\tan x} + \frac{\tan^2\!x}{\tan x}\;\;=\;\; \cot x + 1 +\tan x

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