Let $M=\left[\begin{array}{cc}{\sin }^4\theta & -1-{\sin }^2\theta \\ 1+{\cos }^2\theta & {\cos }^4\theta \end{array}\right]=\alpha I+\beta M^{-1}$ , Where $\alpha =\alpha \left(\theta \right)$  and $\beta =\beta \left(\theta \right)$  are real numbers, and I is the $2\times 2$ identity matrix.

 If $\alpha *$  is the minimum of the set $\left\{\alpha \left(\theta \right):\theta \in [0,2\pi )\right\}$  and $\beta *$   is the minimum of the set $\left\{\beta \left(\theta \right):\theta \in [0,2\pi )\right\}$. 

Then the value of $\alpha *+\beta *$  is

$\begin{array}{l}M=\left[\begin{array}{cc}{\sin }^4\theta & -1-{\sin }^2\theta \\ 1+{\cos }^2\theta & {\cos }^4\theta \end{array}\right]=\alpha I+\beta M^{-1}\\ \text{ Where }\alpha =\alpha \left(\theta \right)\text{ and }\beta =\beta \left(\theta \right)\text{ are real numbers }\\ \left|M\right|=(\sin \theta \cos \theta {)}^4+\left(1+{\sin }^2\theta \right)\left(1+{\cos }^2\theta \right)\\ =2+{\sin }^2\theta {\cos }^2\theta +{\sin }^4\theta {\cos }^4\theta \end{array}$

$\sin ce M=\alpha I+\beta M^{-1}$

$\begin{array}{l}\left[\begin{array}{cc}{\sin }^4\theta & -\left(1+{\sin }^2\theta \right)\\ 1+{\cos }^2\theta & {\cos }^4\theta \end{array}\right]=\left[\begin{array}{cc}\alpha & 0\\ 0& \alpha \end{array}\right]+\frac{\beta }{\left|M\right|}\left[\begin{array}{cc}{\cos }^4\theta & 1+{\sin }^2\theta \\ -1-{\cos }^2\theta & {\sin }^4\theta \end{array}\right]\\ \text{ Comparing on both side }\\ {\sin }^4\theta =\alpha +\frac{\beta }{\left|M\right|}{\cos }^4\theta \\ -1-{\sin }^2\theta =\frac{\beta }{\left|M\right|}\left(1+{\sin }^2\theta \right)\Rightarrow \beta =-\left|M\right| =-\left[2+{\sin }^2\theta {\cos }^2\theta +{\sin }^4\theta {\cos }^4\theta \right]\end{array}$

 $therefore {\sin }^4\theta =\alpha -{\cos }^4\theta \Rightarrow \alpha =\alpha \left(\theta \right)={\sin }^4\theta +{\cos }^4\theta =1-2{\sin }^2\theta {\cos }^2\theta =1-\frac{1}{2}{\sin }^{2}2\theta$

$\begin{array}{l}{\alpha }^{*}=\text{ minimum of }\alpha \left(\theta \right)=1-\frac{1}{2}\left(1\right)=\frac{1}{2}\\ now \beta * =Minimum of \beta \left(\theta \right)=-\left[2+{\sin }^2\theta {\cos }^2\theta +{\sin }^4\theta {\cos }^4\theta \right]=-\left[{\left({\sin }^2\theta {\cos }^2\theta +\frac{1}{2}\right)}^2+\frac{7}{4}\right]=-\left[{\left(\frac{1}{4}{\sin }^{2}2\theta +\frac{1}{2}\right)}^2+\frac{7}{4}\right]\\ {\beta }^{*}=-\left[{\left(\frac{1}{4}+\frac{1}{2}\right)}^2+\frac{7}{4}\right]=-\frac{37}{16} when \theta =n\mathrm{\pi }+\frac{\mathrm{\pi }}{4}\\ {\alpha }^{*}+\beta *=\frac{-37}{16}+\frac{1}{2}=\frac{-37+8}{16}=\frac{-29}{16}\end{array}$