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Math Help - identity

  1. #1
    Junior Member
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    identity

    i don't know how to do this identity:

    tan ^4-sec^4=...

    can you help me? i hope not to disturb you..but i'm reviewing 2 chapters made so far...and i truly need your help.
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  2. #2
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    Hello, alessandromangione!

    You don't have an identity.
    The best we can do is simplify the expression.
    . . (And how far should we go?)


    Simplify:. . \tan^4\!x -\sec^4\!x

    \tan^4\!x - \sec^4\!x \;=\; -(\sec^4\!x - \tan^2\!x)

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

    . . . . . . . . . =\; -(\sec^2\!x + \tan^2\!x)


    . . . . . . . . . =\;\begin{Bmatrix}-(\tan^2\!x + 1 + \tan^2\!x) &=& -2\tan^2\!x - 1 \\ \\<br />
-(\sec^2\!x + \sec^2\!x - 1) &=& -2\sec^2\!x + 1 \\ \\<br />
-\left(\dfrac{1}{\cos^2\!x} + \dfrac{\sin^2\!x}{\cos^2\!x}\right) &=& -\dfrac{1 + \sin^2\!x}{\cos^2\!x}<br /> <br />
 \end{Bmatrix}


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  3. #3
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    i would ike that u help me on doing these ones too...

    sec^4 - 2 sec^2tan^2+ tan^4=

    sec sin / tan -1

    use a right triangle to write the expression as an algebraic expression. Assume that v is positive and in the domainof the given inverse trigonometric function.

    8)cos ( inverse sin v )

    9) sin (inverse sec square root of v^2+9/v) . The possible choices are :

    a)v square root of v^2 +3/v^2+3
    b) v suqare root of v^2/2
    c) 3 square root of v^2 + 9 / v^2 +9
    d) square root of v^2 +3/v^2+3

    find the exact value of
    inverse cos ( cos (-3 radian / 5) )

    i have to do a test to catch up and raise my grades...really that you guys help me!
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  4. #4
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    Dec 2009
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    Quote Originally Posted by Soroban View Post
    Hello, alessandromangione!

    You don't have an identity.
    The best we can do is simplify the expression.
    . . (And how far should we go?)



    \tan^4\!x - \sec^4\!x \;=\; -(\sec^4\!x - \tan^2\!x)

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

    . . . . . . . . . =\; -(\sec^2\!x + \tan^2\!x)


    . . . . . . . . . =\;\begin{Bmatrix}-(\tan^2\!x + 1 + \tan^2\!x) &=& -2\tan^2\!x - 1 \\ \\<br />
-(\sec^2\!x + \sec^2\!x - 1) &=& -2\sec^2\!x + 1 \\ \\<br />
-\left(\dfrac{1}{\cos^2\!x} + \dfrac{\sin^2\!x}{\cos^2\!x}\right) &=& -\dfrac{1 + \sin^2\!x}{\cos^2\!x}<br /> <br />
 \end{Bmatrix}


    thank you dude!!
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