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

We have, $\mathrm{A}={\sin }^2\mathrm{\theta }+{\cos }^4\mathrm{\theta }$

$={\sin }^2\mathrm{\theta }+{\cos }^2{\mathrm{\theta cos}}^2\mathrm{\theta }$

$\le {\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }\left(\text{ since, }{\cos }^2\mathrm{\theta }\le 1\right)$

$\Rightarrow {\sin }^2\mathrm{\theta }+{\cos }^4\mathrm{\theta }\le 1\Rightarrow \mathrm{A}\le 1$

Again, $\begin{array}{r}{\sin }^2\mathrm{\theta }+{\cos }^4\mathrm{\theta }=1-{\cos }^2\mathrm{\theta }+{\cos }^4\mathrm{\theta }\\ ={\cos }^4\mathrm{\theta }-{\cos }^2\mathrm{\theta }+1\\ ={\left({\cos }^2\mathrm{\theta }-\frac{1}{2}\right)}^2+\frac{3}{4}\ge \frac{3}{4}\end{array}$

Hence, $\frac{3}{4}\le \mathrm{A}\le 1$