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Math Help - Prove the following trigonometrical identity-

  1. #1
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    Prove the following trigonometrical identity-

    (1- tan x + sec x)/(1+tan x + sec x) = cos x/(1+sin x)
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  2. #2
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    Hii

    quick question, are you allowed to manipulate the right side of the equation? Some teachers donīt appreciate that..
    What have you tried so far?
    Have you tried writing tan(x)= \frac{sin(x)}{cos(x)}
    and sec(x) = \frac{1}{cos(x)}
    Doing so will easily give you a common denominator on the left side of the = sign...
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  3. #3
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    well i'm afraid i've never tried that....if u know some other method to get this done then plz try that out...
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  4. #4
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    Hello, snigdha!

    If you know this identity: . \sec^2\!x - \tan^2\!x \:=\:1, here's another approach.



    Prove: / \frac{1- \tan x + \sec x)}{1+\tan x + \sec x} \:=\: \frac{\cos x}{1+\sin x}
    We have: . \frac{1 + \sec x - \tan x}{1 + \sec x +\tan x}


    Multiply by \frac{\sec x + \tan x}{\sec x + \tan x}\!:\qquad \frac{\sec x + \tan x}{\sec x + \tan x} \cdot\frac{1 + \sec x - \tan x}{1 + \sec x + \tan x} .  \;\;=\;\; \frac{(\sec x + \tan x) + (\sec x + \tan x)(\sec x - \tan x)}{(\sec x + \tan x)(1 + \sec x + \tan x)}

    . . . . . =\; \frac{(\sec x + \tan x) + \overbrace{(\sec^2\!x - \tan^2\!x)}^{\text{This is 1}}}{(\sec x+\tan x)(1 + \sec x +\tan x)} . =\;\;\frac{\overbrace{\sec x + \tan x + 1}^{\text{These cancel}}}{(\sec x + \tan x)\underbrace{(1 + \sec x +\tan x)}_{\text{These cancel}}}

    . . . . . =\; \frac{1}{\sec x + \tan x} \;\;=\;\;\frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}}


    Multiply by \frac{\cos x}{\cos x}\!:\qquad \frac{\cos x}{\cos x}\cdot \frac{1}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} \;=\; \frac{\cos x}{1 + \sin x}

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  5. #5
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    Hello snigdha
    Quote Originally Posted by snigdha View Post
    (1- tan x + sec x)/(1+tan x + sec x) = cos x/(1+sin x)
    ... and here's an approach that just uses sines and cosines.

    \frac{1-\tan x +\sec x}{1+\tan x + \sec x}=\frac{1-\tan x +\sec x}{1+\tan x + \sec x}\times\frac{\cos^2x}{\cos^2x}
    =\frac{\cos x(\cos x-\sin x +1)}{\cos^2x+\cos x(\sin x + 1)}

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

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

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

    =\frac{\cos x}{1+\sin x}
    Grandad

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  6. #6
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    Soroban and Grandad did it again wih EASE, wow!



    !
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  7. #7
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    Quote Originally Posted by pacman View Post
    Soroban and Grandad did it again wih EASE, wow!



    !
    ... That's what comes of a lifetime of useless activity!

    Grandad
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