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Math Help - Identities!

  1. #1
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    Exclamation Identities!

    I need some guidance on proving these identitles. Any help would be appreciated.

    1)

    ((sec x+1)(sec x-1))/(2+(tan x+1)(tan x-1)) = sin^2 x

    2)

    sec x + csc x = (1+tan x/sin x)

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  2. #2
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    Quote Originally Posted by anna_sims View Post

    2)

    sec x + csc x = (1+tan x/sin x)
    Work on the right side,
    \frac{1+\tan x}{\sin x}

    \frac{1}{\sin x}+\frac{\tan x}{\sin x}

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

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

    \csc x+\sec x
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  3. #3
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    Quote Originally Posted by anna_sims View Post
    1)

    ((sec x+1)(sec x-1))/(2+(tan x+1)(tan x-1)) = sin^2 x
    Well first of all:
    sin^2x + cos^2x = 1

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

    So
    tan^2x + 1 = sec^2x

    Now to the LHS of your expression:
    \frac{(secx + 1)(secx - 1)}{2 + (tanx + 1)(tanx - 1)} Mulitplying out the numerator and denominator:

    \frac{sec^2x -1}{2 + tan^2x - 1}

    \frac{tan^2x}{tan^2 + 1} <-- Using the tan^2 relation in the numerator.

    \frac{tan^2x}{sec^2x} <-- Using the tan^2 relation in the denominator.

    Now convert everything into sines and cosines:
    \frac{ \frac{sin^2x}{cos^2x} }{ \frac{1}{cos^2x} }

    Multiply top and bottom by cos^2x:
    \frac{ \frac{sin^2x}{cos^2x} }{ \frac{1}{cos^2x} } \cdot \frac{cos^2x}{cos^2x} = \frac{sin^2x}{1} = sin^2x

    -Dan
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