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Thread: Another proving identities problem

  1. #1
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    Another proving identities problem

    I've stumbled across another problem that I'm having a hard time solving.

    Prove that:


    $\displaystyle \frac{\ tan(x) - \ sin(x)}{\sin^3(x)}=\frac{1}{\cos(x) + \ cos^2(x)}$


    Help appreciated as always!
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  2. #2
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    Quote Originally Posted by Croton View Post
    I've stumbled across another problem that I'm having a hard time solving.

    Prove that:


    $\displaystyle \frac{\ tan(x) - \ sin(x)}{\sin^3(x)}=\frac{1}{\cos(x) + \ cos^2(x)}$


    Help appreciated as always!
    $\displaystyle \frac{\ tan(x) - \ sin(x)}{\sin^3(x)} = $$\displaystyle \frac{\sin x - \sin x \cos x}{\cos x \sin x \sin^2 x} = \frac{1 - \cos x}{\cos x \sin^2 x}$$\displaystyle \times \frac{1 + \cos x}{1 + \cos x} = \frac{\sin^2 x}{\sin^2 x \cos x (1 + \cos x )} = \frac{1}{\cos x + \cos^2 x}$
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  3. #3
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    Quote Originally Posted by Croton View Post
    I've stumbled across another problem that I'm having a hard time solving.

    Prove that:


    $\displaystyle \frac{\ tan(x) - \ sin(x)}{\sin^3(x)}=\frac{1}{\cos(x) + \ cos^2(x)}$


    Help appreciated as always!

    $\displaystyle \frac{\ tan(x) - \ sin(x)}{\sin^3(x)}$


    $\displaystyle \frac{ \frac{\ sin(x)}{\ cos(x)} - \ sin(x)}{\sin^3(x)}$

    $\displaystyle \frac{\ sin(x)(\frac{1}{\ cos(x)} - 1)}{\sin^3(x)}$

    cancel sin(x)

    $\displaystyle \frac{(\frac{1}{\ cos(x)} - 1)}{\sin^2(x)}$

    Code:
    $\displaystyle \sin^2(x)=1-\ cos(x)$


    $\displaystyle \frac{\frac{1-\cos(x)}{\ cos(x)}}{1-\cos^2(x)}$

    Code:
    remember $\displaystyle (a^2-b^2)=(a-b)(a+b)$
    so, $\displaystyle 1-\cos^2(x)=(1-\cos(x))(1+\cos(x))$
    and
     cancel $\displaystyle 1-cos(x)$







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  4. #4
    MHF Contributor Calculus26's Avatar
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    See attachment
    Attached Thumbnails Attached Thumbnails Another proving identities problem-trigident.jpg  
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  5. #5
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    That helped a lot, thanks!
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