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Prove $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$

So, LS= $$\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}$$ $$\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\right)-\left(\dfrac{1}{\tan\theta}\cdot \dfrac{\sin\theta}{1}\right)$$ $$\dfrac{\tan\theta}{\sin\theta}-\dfrac{\sin\theta}{\tan\theta}$$ Now, considering the fact that I must have a common denominator to subtract, would this be correct:
$$\dfrac{\tan^2\theta}{\tan\theta\sin\theta}-\dfrac{\sin^2\theta}{\tan\theta\sin\theta}\Rightarrow \dfrac{\tan^2\theta-\sin^2\theta}{\tan\theta\sin\theta}$$ I feel like I'm close to the answer because the denominator is the RS of the OP. Please help. Do not give me the answer.

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$$\frac{\tan\theta}{\sin\theta}-\frac{\sin\theta}{\tan\theta}=\frac1{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}=\dots$$

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    $\begingroup$ $=\dfrac{\sin^2\theta}{\cos\theta}\Rightarrow \tan\theta\cdot \dfrac{\sin\theta}{\cos\theta}$ $\endgroup$ – Austin Broussard Aug 21 '12 at 21:25
  • $\begingroup$ @Austin: Yes, and from there? $\endgroup$ – Brian M. Scott Aug 21 '12 at 21:26
  • $\begingroup$ Maybe... $\dfrac{\tan\theta\sin\theta}{\cos\theta}$??? It looks like there are a few ways to go.. $\endgroup$ – Austin Broussard Aug 21 '12 at 21:27
  • $\begingroup$ @Austin: No, $\frac{\sin^2\theta}{\cos\theta}=\frac{\sin\theta}{\cos\theta}\cdot\sin\theta= \tan\theta\sin\theta$. I’ve just used $\frac{\sin\theta}{\cos\theta}= \tan\theta$ again, the same thing that I used twice in the very first step of my answer. $\endgroup$ – Brian M. Scott Aug 21 '12 at 21:31
  • $\begingroup$ Austin, you're right that it's equal to $\cfrac{\sin^2\theta}{\cos\theta}$. That isn't the same thing as $\tan\theta\cdot\cfrac{\sin\theta}{\cos\theta}$, though. $\endgroup$ – Cameron Buie Aug 21 '12 at 21:31
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Brian's answer is short, sweet, and correct. But you might wonder if you can finish it from where you left off. You can.

  1. $\dfrac{\tan^2 \theta - \sin ^2 \theta}{\sin \theta \tan \theta}$
  2. Expand everything into $\sin$ and $\cos$ and simplify into a fraction. Along the way, we get $\dfrac{(\sin^2 \theta - \sin^2 \theta \cos^2 \theta)\cos \theta}{\cos^2 \theta\sin^2 \theta} = \dfrac{\sin^2 \theta \cos \theta(1 - \cos^2 \theta)}{\sin^2 \theta \cos^2 \theta} = \dfrac{(1 - \cos^2 \theta)}{\cos \theta}$
  3. Remember that $\sin^2 \theta + \cos^2 \theta = 1$, and use this to end with the same result of Brian's answer: $\dfrac{\sin^2 \theta}{\cos \theta}$.
  4. Finish it from there.
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Or you could:

$$ \dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta} = \frac{\sin \theta}{\sin \theta \cos\theta} - \frac{\cos\theta\sin\theta}{\sin\theta}. $$ Then cancel terms. Then common denominator. Then...

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It's a great idea to use the reciprocal identities, here. It's also often a good idea to use the quotient identity $\tan\theta=\cfrac{\sin\theta}{\cos\theta}$. That way, we get everything into sines and cosines.

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$$\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\frac{\csc^2\theta-\cot^2\theta}{\cot\theta.\csc\theta}=\frac{1}{\cot\theta\csc\theta}=\frac{1}{\cot\theta}.\frac{1}{\csc\theta}=\tan\theta.\sin\theta$$

I have used the identity $\csc^2\theta-\cot^2\theta=1$ here.

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Sometimes it's much easier to check an identity by transforming it into a sequence of equivalent ones than converting one side into the other. The given identity $$\begin{equation*} \frac{\csc \theta }{\cot \theta }-\frac{\cot \theta }{\csc \theta }=\tan \theta \sin \theta \tag{A} \end{equation*}$$

is equivalent to this one reducing the LHS to a common denominator and using the relations $\tan \theta=1/\cot \theta$, $\sin \theta=1/\csc \theta$ on the RHS $$\begin{eqnarray*} \frac{\csc ^{2}\theta -\cot ^{2}\theta }{\cot \theta \csc \theta } =\frac{1 }{\cot \theta \csc \theta } \Leftrightarrow \csc ^{2}\theta -\cot ^{2}\theta =1. \tag{B} \end{eqnarray*}$$

Since the second identity is equivalent to the Pythagorean identity $$\begin{eqnarray*} \sin ^{2}\theta +\cos ^{2}\theta =1 \Leftrightarrow 1+\cot ^{2}\theta =\csc ^{2}\theta \tag{C} \end{eqnarray*}$$

identity $(\text{A})$ holds.

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