# Thread: Prove

1. ## Prove

Prove that $\displaystyle \frac{1+cos\theta+sin\theta}{1-cos\theta+sin\theta}\equiv\frac{1+cos\theta}{sin\t heta}$
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2. $\displaystyle \frac{1+cos\theta+sin\theta}{1-cos\theta+sin\theta}$

\displaystyle \begin{aligned} &= \frac{1 + \sin \theta + \cos \theta}{1 + \sin \theta - \cos \theta} \\\\ &= \frac{[(1 + \sin \theta) + \cos \theta][(1 + \sin \theta) + \cos \theta]}{[(1 + \sin \theta) - \cos \theta][(1 + \sin \theta) + \cos \theta]} \\\\ &= \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} \end{aligned}

\displaystyle \begin{aligned} &= \frac{2 + 2\sin \theta + 2\cos \theta + 2\cos \theta \sin \theta}{2\sin\theta + 2\sin^2 \theta} \\\\ &= \frac{1(2 + 2\sin \theta) + \cos \theta(2 + 2\sin \theta)}{\sin\theta(2 + 2\sin \theta)} \end{aligned}

\displaystyle \begin{aligned} &= \frac{(1 + \cos \theta)(2 + 2\sin \theta)}{\sin\theta(2 + 2\sin \theta)} \\\\ &= \frac{1 + \cos \theta}{\sin\theta} \end{aligned}

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