# Prove the identity $\csc2\theta=\frac{\sec\theta\csc\theta}{2}$

Prove the identity $$\csc2\theta=\frac{\sec\theta\csc\theta}{2}$$

I've started by using a double angle identity, but I'm not sure how to continue from here or if this is right approach.

$$\csc2\theta=\frac{1}{\sin2\theta}=\frac{1}{2\sin\theta\cos\theta}$$

• Your left-hand side should be $\csc (2\theta)$ – paw88789 Dec 4 '18 at 19:20

## 3 Answers

What you did is correct. Now, you do:$$\frac1{2\sin\theta\cos\theta}=\frac{\frac1{\cos\theta}\times\frac1{\sin\theta}}2=\frac{\sec\theta\csc\theta}2.$$

Your approach is right, you just need to do the last step to get the desired result: What are $$\frac{1}{\sin(\theta)}$$ and $$\frac{1}{\cos(\theta)}$$?

The thing you did is OK. The question which occurs is do you know how to prove $$\sin (2x) = 2\sin x \cos x$$

(Hint: use adition theorem for function $$\sin$$)