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Math Help - Prove

  1. #1
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    Prove

    Prove that \frac{1+cos\theta+sin\theta}{1-cos\theta+sin\theta}\equiv\frac{1+cos\theta}{sin\t  heta}
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  2. #2
    Super Member
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    \frac{1+cos\theta+sin\theta}{1-cos\theta+sin\theta}

    \begin{aligned}<br />
&= \frac{1 + \sin \theta + \cos \theta}{1 + \sin \theta - \cos \theta} \\\\<br />
&= \frac{[(1 + \sin \theta) + \cos \theta][(1 + \sin \theta) + \cos \theta]}{[(1 + \sin \theta) - \cos \theta][(1 + \sin \theta) + \cos \theta]} \\\\<br />
&= \frac{1 + 2\sin \theta + \sin^2 \theta + 2\cos \theta + 2\cos \theta \sin \theta + \cos^2 \theta}{1 + 2\sin\theta + \sin^2 \theta - \cos^2 \theta}<br />
\end{aligned}

    \begin{aligned}<br />
&= \frac{2 + 2\sin \theta + 2\cos \theta + 2\cos \theta \sin \theta}{2\sin\theta + 2\sin^2 \theta} \\\\<br />
&= \frac{1(2 + 2\sin \theta) + \cos \theta(2  + 2\sin \theta)}{\sin\theta(2 + 2\sin \theta)}<br />
\end{aligned}

    \begin{aligned}<br />
&= \frac{(1 + \cos \theta)(2 + 2\sin \theta)}{\sin\theta(2 + 2\sin \theta)} \\\\<br />
&= \frac{1 + \cos \theta}{\sin\theta}<br />
\end{aligned}


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