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!

  • $\begingroup$ What "cross-cancelling"? You are subtracting the fractions, not multiplying them. $\endgroup$ – Arturo Magidin Jul 15 '12 at 0:39

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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  • $\begingroup$ Yes I just realized that. I'm new to LaTeX. And I was trying to figure out how to divide fractions. $\endgroup$ – Austin Broussard Jul 15 '12 at 0:47
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    $\begingroup$ @Austin: Use \frac, not \dfrac. If you rightclick on my formula, you can ask that it show you the LaTeX code I used to produce it. $\endgroup$ – Arturo Magidin Jul 15 '12 at 0:57
  • $\begingroup$ Thank you so much. You have been helpful today. And I'm re-editing my post with what I feel is correct. Stay posted! $\endgroup$ – Austin Broussard Jul 15 '12 at 1:00
  • $\begingroup$ Is the right side of my equation good? Or is there anything you have to say about that so I can look for equality on both sides. $\endgroup$ – Austin Broussard Jul 15 '12 at 1:11
  • $\begingroup$ @Austin: Your computations with $\tan\theta\sin\theta$ are correct. So you want to take the left hand side and change everything into sines and cosines, like I did, and simplify until you have a single fraction (instead of a compound fraction which is what I have up to where I developed it). You should be able to get it to be exactly the same as what you got for the right hand side. $\endgroup$ – Arturo Magidin Jul 15 '12 at 1:13

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