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Math Help - 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:


    \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:


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


    Help appreciated as always!
    \frac{\ tan(x) - \ sin(x)}{\sin^3(x)} = \frac{\sin x - \sin x \cos x}{\cos x \sin x \sin^2 x} = \frac{1 - \cos x}{\cos x \sin^2 x}  \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:


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


    Help appreciated as always!

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


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

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

    cancel sin(x)

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

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


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

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