0
$\begingroup$

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.

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

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.