Trigonometric Identities help please $$  \frac {\sec \theta}{  \csc \theta - \cot \theta }  - \frac {  \sec \theta }{  \csc \theta + \cot \theta }  =  2\csc \theta $$
I really have no idea how to verify. I try then but can't make sense of it. Thanks in advance for any help.
 A: Hint
Get a common denominator on left hand side and then use the fact that $\csc^2\theta-\cot^2\theta=1$ and $\cot\theta=\frac{\cos\theta}{\sin\theta}$ and $\sec\theta=\frac{1}{\cos\theta}$
A: $$\frac {\sec \theta}{  \csc \theta - \cot \theta }  - \frac {  \sec \theta }{  \csc \theta + \cot \theta }  =  2\csc \theta$$
Simplifying the LHS,
$$\frac{\sec \theta(\csc \theta +\cot \theta) -\sec \theta (\csc \theta -\cot \theta}{\csc^2 \theta - \cot^2 \theta} $$   [ Cross multiplying to make the denominators same.]
$$\implies \frac{\sec \theta (\csc \theta + \cot \theta -1(\csc \theta - \cot \theta)}{1}  $$         [Taking $$\sec \theta $$common and equating$$ \csc^2 - \cot^2 \theta = 1$$]
$$\implies \frac{ \sec \theta(\csc \theta + \cot \theta -\csc \theta + \cot \theta)}{1}$$
$$\implies \frac \sec \theta(\cot \theta + \cot \theta)$$
$$\implies \sec \theta *  2 * \cot \theta$$
$$\implies \frac{1}{\cos \theta} * 2 * \frac{\cos \theta}{\sin \theta} $$   [Equating $$\sec \theta = \frac{1}{\cos \theta}  and  \cot \theta = \frac{\cos \theta}{\sin \theta}$$]
$$\implies 2 * \frac{1}{\sin \theta}$$
$$\implies 2 * \csc \theta $$  [As $$\frac{1}{\sin \theta} = \csc \theta$$]
LHS = RHS, hence verified
Please note that there are many other different ways to verify the equation. Hope it helps!
