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Math Help - Verifying Identities?

  1. #1
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    Verifying Identities?

    I have two questions that I am getting nowhere on:

    1. Verify the identity: cotx+tanx=cscxsecx
    I thought of working on the right side, but I came up with (1/tanx)+(sinx/cosx) and didn't know where to go from there.

    2. Verify the identity: (15sinx-4cos^2x)/(4+sinx)=4sinx-1
    I don't even know where to begin on this one!

    Any help is much appreciated, and since I need to show work, a brief explanation is welcomed. Thanks in advance!!
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  2. #2
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    Quote Originally Posted by captaintoast87 View Post
    I have two questions that I am getting nowhere on:

    1. Verify the identity: cotx+tanx=cscxsecx
    I thought of working on the right side, but I came up with (1/tanx)+(sinx/cosx) and didn't know where to go from there.

    2. Verify the identity: (15sinx-4cos^2x)/(4+sinx)=4sinx-1
    I don't even know where to begin on this one!

    Any help is much appreciated, and since I need to show work, a brief explanation is welcomed. Thanks in advance!!
    1. That's because there is nothing more to do, just change them into the LHS using the rules below

    tan(x) = \frac{sin(x)}{cos(x)} and cot(x) = \frac{1}{tan(x)}

    2. cos^2(x) = 1-sin^2(x) so -4cos^2(x) = -4(1-sin^2(x)) = -4 + 4sin^2(x)

    <br />
\frac{15sin(x)+4sin^2(x)-4}{4+sin(x)}

    The numerator factorises to (4sin(x)-1)(sin(x)+4)

    \frac{(4sin(x)-1)(sin(x)+4)}{sin(x)+4} = 4sin(x)-1
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  3. #3
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    Quote Originally Posted by captaintoast87 View Post
    I have two questions that I am getting nowhere on:

    1. Verify the identity: cotx+tanx=cscxsecx
    I thought of working on the right side, but I came up with (1/tanx)+(sinx/cosx) and didn't know where to go from there.

    2. Verify the identity: (15sinx-4cos^2x)/(4+sinx)=4sinx-1
    I don't even know where to begin on this one!

    Any help is much appreciated, and since I need to show work, a brief explanation is welcomed. Thanks in advance!!
    Hi captaintoast87,

    Here's your first one:

    \cot x + \tan x = \csc x sec x

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

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

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

    \frac{1}{\sin x} \cdot \frac{1}{\cos x}=

    \csc x \sec x
    Last edited by masters; March 31st 2009 at 12:37 PM.
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by masters View Post
    Hi captaintoast87,

    Here's your first one:

    \cot x + \tan x = \csc x sec x

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

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

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

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

    \csc x \sec x
    How did you get from \frac{1}{\sin x \cos x}= to
    \frac{1}{\sin x}+\frac{1}{\cos x}=
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by e^(i*pi) View Post
    How did you get from \frac{1}{\sin x \cos x}= to
    \frac{1}{\sin x}+\frac{1}{\cos x}=
    Rookie mistake. Typo. Should've been a product, not a sum.

    Fixed it. Thanks!
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