# Verifying some trigonometric identities: $\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$ [duplicate]

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!

• What "cross-cancelling"? You are subtracting the fractions, not multiplying them. Jul 15, 2012 at 0:39

\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?
• @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. Jul 15, 2012 at 0:57
• @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. Jul 15, 2012 at 1:13