$$\frac {\sec \theta}{ \csc \theta - \cot \theta } - \frac { \sec \theta }{ \csc \theta + \cot \theta } = 2\csc \theta$$

I really have no idea how to verify. I try then but can't make sense of it. Thanks in advance for any help.

• – Shaun May 14 '15 at 17:21
• Hint : $\csc^2 \theta - \cot^2 \theta =1$ $\implies$ $\csc \theta - \cot \theta = \frac{1}{\csc \theta + \cot \theta}$ – Mann May 14 '15 at 17:21
• I'm tempted to downvote for not even proofreading the subject title. Please move your left hand one key to the right. – Matthew Leingang May 14 '15 at 17:26

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 $\sec\theta=\frac{1}{\cos\theta}$

• Also $\csc\theta = \frac{1}{\sin\theta}$, of course. – user26486 May 14 '15 at 18:11
• Absolutely :) ${}$ – homegrown May 14 '15 at 18:25

$$\frac {\sec \theta}{ \csc \theta - \cot \theta } - \frac { \sec \theta }{ \csc \theta + \cot \theta } = 2\csc \theta$$

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!