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Math Help - Identity crisis

  1. #1
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    Exclamation Identity crisis

    Heres the problem, Prove the identity. I can't for the life of me to figure it out.

    Cosx+sinxtanx = cscx
    sinxsecx
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by nightrider456 View Post
    Heres the problem, Prove the identity. I can't for the life of me to figure it out.

    Cosx+sinxtanx = cscx
    sinxsecx

    \dfrac{\cos x + \sin x \tan x }{\sin x \sec x}


    \frac{\cos x + \sin x \cdot \dfrac{\sin x}{\cos x}}{\sin x \dfrac{1}{\cos x }}


    \dfrac{\dfrac{\cos ^2 x + \sin ^2 x}{\cos x}}{\dfrac{\sin x}{\cos x}}


    \frac{\cos ^2 x + \sin ^2 x }{\sin x } = \frac{1}{\sin x} = \csc x
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  3. #3
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    Quote Originally Posted by nightrider456 View Post
    Heres the problem, Prove the identity. I can't for the life of me to figure it out.

    Cosx+sinxtanx = cscx
    sinxsecx
    \frac{\cos{x} + \sin{x}\tan{x}}{\sin{x}\sec{x}} = \frac{\cos{x}}{\sin{x}\sec{x}} + \frac{\sin{x}\tan{x}}{\sin{x}\sec{x}}

     = \frac{\cos^2{x}}{\sin{x}} + \frac{\tan{x}}{\sec{x}}

     = \frac{\cos^2{x}}{\sin{x}} + \frac{\frac{\sin{x}}{\cos{x}}}{\frac{1}{\cos{x}}}

     = \frac{\cos^2{x}}{\sin{x}} + \sin{x}

     = \frac{\cos^2{x}}{\sin{x}} + \frac{\sin^2{x}}{\sin{x}}

     = \frac{\cos^2{x} + \sin^2{x}}{\sin{x}}

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

     = \csc{x}.
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