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Thread: Simplifying using trig identities

  1. #1
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    Simplifying using trig identities

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

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

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

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


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

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

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

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

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


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

    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..

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

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

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

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

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