If $\sin x+{\sin }^{2}x=1$, then ${\cos }^{2}x+{\cos }^{4}x=$

Method-I

$\sin x+{\sin }^{2}x=1$

$\Rightarrow \sin x=1-{\sin }^{2}x$

$\Rightarrow \sin x={\cos }^{2}x$

${\cos }^{2}x+{\cos }^{4}x$

$\Rightarrow {\cos }^{2}x+{\left({\cos }^{2}x\right)}^2$

$={\cos }^{2}x+{\left(\sin x\right)}^2$

$=1$

Method-II

$\sin x+{\sin }^{2}x=1$

$\Rightarrow \sin x=1-{\sin }^{2}x$

$\Rightarrow \sin x={\cos }^{2}x$

Squaring on both sides

$\Rightarrow {\sin }^{2}x={\cos }^{4}x$

$\Rightarrow 1-{\cos }^{2}x={\cos }^{4}x$

$\Rightarrow {\cos }^{2}x+{\cos }^{4}x=1$