# Hard C3 question

1. Jun 4, 2008

### thomas49th

im stuck on the top question on part B

i can do the difference of 2 squares on the LHS but how do i get rid of the cosec²x - cot²x ??

Thanks

2. Jun 4, 2008

### 3lliott

mis-post, oops!

Edit: Well, as I'm here...

$cosec^2\theta - cot^2\theta = 1$
$cosec^2\theta + cot^2\theta = -1$

Well, that's the RHS...

$-cot^4\theta = -cot^2\theta \times cot^2\theta\newline$
$= (1 - cosec^2\theta)(cosec^2\theta - 1)\newline$
$= -1 + cosec^2\theta - cosec^4\theta - cosec^2\theta\newline$

Back to the original question!

$cosec^4\theta - cosec^4\theta - 1 = - 1\newline$
$-1 = -1$

I think that's correct :S

(Hmm, latex is messed up, how do I make a newline?)

Last edited: Jun 4, 2008
3. Jun 4, 2008

### Gib Z

Use the result in part a) like it told you to.

4. Jun 4, 2008

### D H

Staff Emeritus
3lliot, please do not give out answers. We offer free help here at PF, not free answers.