Proving that $\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$ 
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.
 A: $$\frac{\tan\theta}{\sin\theta}-\frac{\sin\theta}{\tan\theta}=\frac1{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}=\dots$$
A: Brian's answer is short, sweet, and correct. But you might wonder if you can finish it from where you left off. You can.


*

*$\dfrac{\tan^2 \theta - \sin ^2 \theta}{\sin \theta \tan \theta}$

*Expand everything into $\sin$ and $\cos$ and simplify into a fraction. Along the way, we get $\dfrac{(\sin^2 \theta - \sin^2 \theta \cos^2 \theta)\cos \theta}{\cos^2 \theta\sin^2 \theta} = \dfrac{\sin^2 \theta \cos \theta(1 - \cos^2 \theta)}{\sin^2 \theta \cos^2 \theta} = \dfrac{(1 - \cos^2 \theta)}{\cos \theta}$

*Remember that $\sin^2 \theta + \cos^2 \theta = 1$, and use this to end with the same result of Brian's answer: $\dfrac{\sin^2 \theta}{\cos \theta}$.

*Finish it from there.

A: 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...
A: 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.
A: 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.
A: $$\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\frac{\csc^2\theta-\cot^2\theta}{\cot\theta.\csc\theta}=\frac{1}{\cot\theta\csc\theta}=\frac{1}{\cot\theta}.\frac{1}{\csc\theta}=\tan\theta.\sin\theta$$
I have used the identity $\csc^2\theta-\cot^2\theta=1$ here.
