If sinx+sin2x=1 then the value of cos2x+cos4x+cot4x−cot2x is?

Given
$\sin x+{\sin }^{2}x=1$
$\Rightarrow \sin x=1-{\sin }^{2}x$
$\Rightarrow \sin x={\cos }^{2}x$
$\Rightarrow {\sin }^{2}x={\cos }^{4}x$
$\Rightarrow 1-{\cos }^{2}x={\cos }^{4}x$
$\Rightarrow {\cos }^{2}x+{\cos }^{4}x=1......\left[1\right]$
Again
$\sin x+{\sin }^{2}x=1$
$\Rightarrow \frac{\sin x}{{\sin }^{2}x}+\frac{{\sin }^{2}x}{{\sin }^{2}x}=\frac{1}{{\sin }^{2}x}$
$\Rightarrow \csc x+1={\csc }^{2}x$
$\Rightarrow \csc x={\csc }^{2}x-1$
$\Rightarrow \csc x={\cot }^{2}x$
$\Rightarrow {\csc }^{2}x={\cot }^{4}x$
$\Rightarrow 1+{\cot }^{2}x={\cot }^{4}x$
$\Rightarrow {\cot }^{4}x-{\cot }^{2}x=1........\left[2\right]$
Adding [1] and [2]
we get
cos2x+cos4x+cot4x−cot2x=1+1=2