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Prove the following:
46. $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$

I got as far as
Right Side: $\tan\theta\sin\theta$ to $\dfrac{\sin\theta}{\cos\theta}\dfrac{\sin\theta}{1}$ and then; $\dfrac{\sin^2\theta}{\cos\theta}$

Left Side: $\begin{align*} \dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta} &= \dfrac{\frac{1}{\sin^2\theta}-{\frac{\cos^2\theta}{\sin^2\theta}}}{\frac{\cos\theta}{\sin\theta}-{\frac{1}{\sin^2\theta}}}\\ &= \dfrac{\frac{1-\cos^2\theta}{\sin^2\theta}}{\frac{\cos\theta}{\sin^2\theta}}\\ &= \dfrac{\frac{\sin^2\theta}{\sin^2\theta}}{\frac{\cos\theta}{\sin^2\theta}}\\ &= \frac{1}{\frac{\cos\theta}{\sin^2\theta}}\\ &= \frac{\sin^2\theta}{\cos\theta} \end{align*}$ Thanks a lot!

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    What "cross-cancelling"? You are subtracting the fractions, not multiplying them.2012-07-15

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There is no "cross cancelling". You are subtracting the fractions, not multiplying them.

$\begin{align*} \frac{\csc\theta}{\cot\theta} - \frac{\cot\theta}{\csc\theta} & = \frac{\csc^2\theta - \cot^2\theta}{\cot\theta\csc\theta}\\ &= \frac{\quad\frac{1}{\sin^2\theta} - \frac{\cos^2\theta}{\sin^2\theta}\quad}{\frac{\cos\theta}{\sin\theta}\frac{1}{\sin\theta}}\\ &= \frac{\quad\frac{1 - \cos^2\theta}{\sin^2\theta}\quad}{\frac{\cos\theta}{\sin^2\theta}}. \end{align*}$ Can you take it from there?

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    @Austin: If it's a new problem, please consider writing a new question. Otherwise, the answer will no longer fit.2012-07-15