# Thread: Verifing these 2 identities?

1. ## Verifing these 2 identities?

I am so confused on how to verify these. Any help is appreciated

its cot squared X
csc[cscX+sin(-X)]= cot^2)(X)

and

its sin & cos to the 4th X and its sin & cos squared X
[Sin^4)(Θ)]-[Cos^4)(Θ)]=[Sin^2)(Θ)]-[Cos^2)(Θ)]

2. Originally Posted by snohosoccer18
I am so confused on how to verify these. Any help is appreciated

its cot squared X
csc[cscX+sin(-X)]= cot^2)(X)
Hi snohosoccer18,

Could it possibly be $\displaystyle \csc x[\csc x + \sin(-x)]=\cot^2(x)$

3. yea that is it. i just don't know how to post that in the text box.

4. Originally Posted by masters
Hi snohosoccer18,

Could it possibly be $\displaystyle \csc x[\csc x + \sin(-x)]=\cot^2(x)$
Ok, here goes.

$\displaystyle \csc x[\csc x + \sin(-x)]=\cot^2(x)$

$\displaystyle \csc x(\csc x - \sin x)=\cot^2 x$

$\displaystyle \csc x\left(\frac{1}{\sin x}-\sin x\right)=\cot^2 x$

$\displaystyle \csc x\left(\frac{1-\sin^2 x}{\sin x}\right)=\cot^2 x$

$\displaystyle \frac{1}{\sin x}\left(\frac{\cos^2 x}{\sin x}\right)=\cot^2 x$

$\displaystyle \frac{\cos^2 x}{\sin^2 x}=\cot^2 x$

$\displaystyle \cot^2 x=\cot^2 x$