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Math Help - Simplifying using trig identities

  1. #1
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    Simplifying using trig identities

    Simplify: \cot(-x) (\cos x) (\tan^2 x + 1)

    I'll show you what I have so far. My problem is I don't really know when I'm finished.

    \cot(-x) (\cos x) (\tan^2 x + 1)
    (\frac{1}{-\tan x)})(\frac{1}{\sec x})(\sec^2 x)
    (\frac{1}{-\tan x})(\sec x)

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


    and I have my final answer as \frac{-\sin x}{\cos^2 x}

    Is this where I should end it, or should I keep going?

    Thanks.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Savior_Self View Post
    Simplify: \cot(-x) (\cos x) (\tan^2 x + 1)

    I'll show you what I have so far. My problem is I don't really know when I'm finished.

    \cot(-x) (\cos x) (\tan^2 x + 1)
    (\frac{1}{-\tan x)})(\frac{1}{\sec x})(\sec^2 x)
    (\frac{1}{-\tan x})(\sec x)

    (\frac{-\sin x}{\cos x})(\frac{1}{\cos x}) NO!!!


    and I have my final answer as \frac{-\sin x}{\cos^2 x}

    Is this where I should end it, or should I keep going?

    Thanks.

    there's a mistake in your work: look at line number 4 of your answer..

    (\frac{1}{-\tan x})(\sec x)

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

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

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

    =-\csc{x}
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  3. #3
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    Ahhhhhh, I'm an idiot. Thanks, harish.
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