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Math Help - Verifying the following identity

  1. #1
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    Verifying the following identity

    Okay hai guys so I'm trying to verify the following identity: \frac{\sin^2\theta+2\cos\theta-1}{\sin^2\theta+3\cos\theta-3}=\frac{1}{1-\sec\theta}

    My attempt:

    \frac{1-\cos^2\theta+2\cos\theta-1}{1-cos^2\theta+3\cos\theta-3}

    \frac{-\cos^2\theta+2\cos\theta}{-cos^2\theta+3\cos\theta-2}

    \frac{-\cos\theta(\cos\theta-2)}{(\cos\theta-1)(\cos\theta-2)}

    -\frac{\cos\theta}{\cos\theta-1}

    Now I'm going to work on the right side

    \frac{1}{1-\sec\theta}

    \frac{1}{1-\frac{1}{\cos\theta}}

    \frac{1}{\frac{\cos\theta-1}{cosx}}

    \frac{cosx}{\cos\theta-1}

    So -\frac{\cos\theta}{\cos\theta-1}=\frac{cosx}{\cos\theta-1}


    What am I doing wrong?
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  2. #2
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    Your factorisation is wrong.

    \frac{-\cos^2{\theta} + 2\cos{\theta}}{-\cos^2{\theta} + 3\cos{\theta} - 2} = \frac{-\cos{\theta}(\cos{\theta} - 2)}{-\cos^2{\theta} + 2\cos{\theta} + \cos{\theta} - 2}

     = \frac{-\cos{\theta}(\cos{\theta} - 2)}{-\cos{\theta}(\cos{\theta} - 2) + 1(\cos{\theta} - 2)}

     = \frac{-\cos{\theta}(\cos{\theta} - 2)}{(-\cos{\theta} + 1)(\cos{\theta} - 2)}

     = \frac{-\cos{\theta}}{-\cos{\theta} + 1}

     = \frac{\cos{\theta}}{\cos{\theta} - 1}
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  3. #3
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    Quote Originally Posted by iamanoobatmath View Post
    Okay hai guys so I'm trying to verify the following identity: \frac{\sin^2\theta+2\cos\theta-1}{\sin^2\theta+3\cos\theta-3}=\frac{1}{1-\sec\theta}

    My attempt:

    \frac{1-\cos^2\theta+2\cos\theta-1}{1-cos^2\theta+3\cos\theta-3}

    \frac{-\cos^2\theta+2\cos\theta}{-cos^2\theta+3\cos\theta-2}

    \frac{-\cos\theta(\cos\theta-2)}{(\cos\theta-1)(\cos\theta-2)}

    -\frac{\cos\theta}{\cos\theta-1} ***

    Now I'm going to work on the right side

    \frac{1}{1-\sec\theta}

    \frac{1}{1-\frac{1}{\cos\theta}}

    \frac{1}{\frac{\cos\theta-1}{cosx}}

    \frac{cosx}{\cos\theta-1}

    So -\frac{\cos\theta}{\cos\theta-1}=\frac{cosx}{\cos\theta-1}


    What am I doing wrong?
    hi

    your factorisation part is wrong ,

    continuing from ***, where the negative sign shouldn't be there,

    \frac{\cos \theta}{\cos \theta-1}

    \frac{\frac{1}{\sec \theta}}{\frac{1}{\sec \theta}-1}

    Then combine the fractions is the denominator , and see where you can go from there .
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