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Math Help - Quick trig problem

  1. #1
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    Quick trig problem

    Truth is, I haven't simplified trig problems in a while. I was wondering if you guys could help me out.

    Here's the problem:

    Write the trigonometric expression in terms of sine and cosine, and then simplify.

    tan(x)/(sec(x)-cos(x))

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by elven06 View Post
    Truth is, I haven't simplified trig problems in a while. I was wondering if you guys could help me out.

    Here's the problem:

    Write the trigonometric expression in terms of sine and cosine, and then simplify.

    tan(x)/(sec(x)-cos(x))

    Thanks in advance!
    note that \tan{x} = \frac{\sin{x}}{\cos{x}}

    and \sec{x} = \frac{1}{\cos{x}}

    proceed with your simplification ...
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  3. #3
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    I tried, but I came out with:

    sin(x)-(2/1+cos(x))

    I'm not sure where I'm going wrong. Thanks for the quick reply.
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  4. #4
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    Quote Originally Posted by elven06 View Post
    I tried, but I came out with:

    sin(x)-(2/1+cos(x))

    I'm not sure where I'm going wrong. Thanks for the quick reply.
    \frac{\sin(x)}{\cos(x)} \times \frac{1}{\frac{1}{\cos(x)}-\cos(x)}


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

    \therefore \: \: \frac{1}{\sec(x)-\cos(x)} = \frac{\cos(x)}{\sin^2(x)}

    \frac{\sin(x)}{cos(x)} \times \frac{\cos(x)}{\sin^2(x)} = \frac{1}{\sin(x)} = \csc(x)
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  5. #5
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    Hello, elven06!

    A slightly different approach . . .


    Write the trigonometric expression in terms of sine and cosine, and then simplify.

    . . \frac{\tan x}{\sec x-\cos x}
    We have: . \frac{\dfrac{\sin x}{\cos x}}{\dfrac{1}{\cos x} - \cos x}

    Multiply by \frac{\cos x}{\cos x}\!:\quad \frac{\cos x\left(\dfrac{\sin x}{\cos x}\right) }{\cos x\left(\dfrac{1}{\cos x} - \cos x\right)} \;=\;\frac{\sin x}{1-\cos^2\!x} \;=\;\frac{\sin x}{\sin^2\!x} \;=\;\frac{1}{\sin x} \;=\;\csc x

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  6. #6
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    Thanks guys, that was it. I had put 1/sinx, but it wanted csc...
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