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.

Solution 1:

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

Solution 2:

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.