Given $A={\sin }^2\theta +{\cos }^4\theta ,$  then for all real values of $\theta$

Given, $A={\sin }^2\theta +{\left(1-{\sin }^2\theta \right)}^2$ $\Rightarrow A={\sin }^4\theta -{\sin }^2\theta +1$ 

$\Rightarrow A={\left({\sin }^2\theta -\frac{1}{2}\right)}^2+\frac{3}{4}$  

Also $A={\sin }^2\theta +{\cos }^4\theta \le {\sin }^2\theta +{\cos }^2\theta \le 1$ 

$\therefore \frac{3}{4}\le A\le 1$