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

Question

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.

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

Answer 1

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

Answer 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 $\frac{1}{\cos(\theta)}$?

Answer 3

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 $)

Answer 4

You are already on the right track. Here is another approach.
GO! $$\begin{aligned}\csc{2\theta}&=\frac{1}{\sin{2\theta}}\\&=\frac{1}{2\sin\theta\cos\theta}\\&=\frac12\csc\theta\sec\theta\\&=\frac{\csc\theta\sec\theta}{2}\end{aligned}$$ I hope this helps.