$A={\sin }^8\theta +{\cos }^{14}\theta ,$then for all values of  $\theta$

$$\begin{gathered} Since 0\le {\sin }^2\theta \le 1 \\ \Rightarrow {\left({\sin }^2\theta \right)}^4\le {\sin }^2\theta \end{gathered}$$

$\Rightarrow {\sin }^8\theta \le {\sin }^2\theta$

$$\begin{gathered} Also 0\le {\cos }^2\theta \le 1 \\ \Rightarrow {\left({\cos }^2\theta \right)}^7\le {\cos }^2\theta \\ \Rightarrow {\cos }^{14}\theta \le {\cos }^2\theta \end{gathered}$$

Now ${\sin }^8\theta +{\cos }^{14}\theta \le {\sin }^2\theta +\cos \theta$

$\Rightarrow A\le 1\&A>0$

$\Rightarrow 0<A\le 1$