Simplify. 2csc^2 y - csc^4 y + cot^4 y

Question

abb0t · Answer

$$2\frac{ 1 }{ \sin^2(y) }-\frac{ 1 }{ \sin^4(y) }+\frac{ \cos^4(x) }{ \sin^4(x) }$$

foolaroundmath · Answer

$\csc^{2}{\theta} = 1 + \cot^{2}\theta$

$\cot^{4}\theta - \csc^{4}\theta = (\cot^{2}\theta+\csc^{2}\theta)(\cot^{2}\theta - \csc^{2}\theta)$
$= -(\cot^{2}\theta + \csc^{2}\theta)$

So our expression simplifies to:
$2\csc^{2}\theta - csc^{2}\theta - \cot^{2}\theta = \csc^{2}\theta -\cot^{2}\theta = 1$

zehanz · Answer

abb0t is right (replace x with y). Now write it as one fraction:$$\frac{ 2\sin^2y }{ \sin^4y }-\frac{ 1 }{ \sin^4y }+\frac{ \cos^4y }{ \sin^4y }=\frac{ 2\sin^2y-1+\cos^4y }{ \sin^4y }$$But it must be possible to simplify further...

zehanz · Answer

@FoolAroundmath:Looks nice! 
I'm from Europe and back here we're not into sec and csc functions as in the USA, so I'll just plough on:$$\frac{ 2\sin^2y-1+\cos^2y \cos^2y }{ \sin^4y }=\frac{ 2\sin^2y-1+(1-\sin^2y)^2 }{ \sin^4y }$$$$=\frac{ 2\sin^2y-1+1-2\sin^2y+\sin^4y }{ \sin^4y }=\frac{ \sin^4y }{ \sin^4y }=1$$