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Math Help - How do I get the final solution?

  1. #1
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    Exclamation How do I get the final solution?

    Hi

    Could someone please show me how to get from the LHS to the RHS???
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  2. #2
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    Quote Originally Posted by Sunyata View Post
    Hi

    Could someone please show me how to get from the LHS to the RHS???
    \frac{atan\theta}{tan\theta-sec\theta+1}=\frac{a\frac{sin\theta}{cos\theta}}{\  frac{sin\theta}{cos\theta}-\frac{1}{cos\theta}+1}

    =\frac{asin\theta}{sin\theta-1+cos\theta}=\frac{asin\left(\frac{2\theta}{2}\rig  ht)}{sin\left(\frac{2\theta}{2}\right)+cos\left(\f  rac{2\theta}{2}\right)-1}

    =\frac{2asin\frac{\theta}{2}cos\frac{\theta}{2}}{2  sin\frac{\theta}{2}cos\frac{\theta}{2}+cos^2\left(  \frac{\theta}{2}\right)-sin^2\left(\frac{\theta}{2}\right)-1}

    =\frac{2asin\frac{\theta}{2}cos\frac{\theta}{2}}{2  sin\frac{\theta}{2}cos\frac{\theta}{2}+\left(1-sin^2\left(\frac{\theta}{2}\right)\right)-sin^2\left(\frac{\theta}{2}\right)-1}

    =\frac{2asin\frac{\theta}{2}cos\frac{\theta}{2}}{2  sin\frac{\theta}{2}cos\frac{\theta}{2}-2sin^2\left(\frac{\theta}{2}\right)}

    =\frac{2acos\frac{\theta}{2}}{2cos\frac{\theta}{2}-2sin\frac{\theta}{2}}

    =\frac{acos\frac{\theta}{2}}{cos\frac{\theta}{2}-sin\frac{\theta}{2}}
    Last edited by Archie Meade; March 23rd 2010 at 05:07 AM. Reason: added parentheses
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