Basic trigonometry

The given expression can be rewritten as,
$sin^2x+cosec^2x+cos^2x+sec^2x+4$
= $4+(sin^2x+cos^2x)+(1+tan^2x)+(1+cot^2x)$
= $7+tan^2x+cot^2x$ ...(1)

Using A.M $\geqslant$ G.M.
$\frac{tan^2x+cot^2x}{2}\geqslant\sqrt[2]{tan^2x.cot^2x}$
$\implies tan^2x+cot^2x\geqslant2$ ...(2)

From (1) and (2),

$(sinx+cosecx)^2+(cosx+secx)^2\geqslant9$

$\begin{aligned} S & = (\sin x + \csc x)^2 + (\cos x + \sec x)^2 \\ & = \left(\sin x + \frac{1}{\sin x} \right)^2 + \left(\cos x + \frac{1}{\cos x} \right)^2 \\ & = \sin^2 x + 2 + \frac{1}{\sin^2 x} + \cos^2 x + 2 + \frac{1}{\cos^2 x} \\ & = 5 + \frac{1}{\sin^2 x} + \frac{1}{\cos^2 x} \\ & = 5 + \frac{1}{\sin^2 x \cos^2 x} \\ & = 5 + \color{#3D99F6}{\frac{4}{\sin^2 2x}} \quad \quad \small \color{#3D99F6}{\text{We note that } \sin^2 2x \ge 1 \quad \Rightarrow \frac{4}{\sin^2 2x} \le 4} \\ & \le 5 + \color{#3D99F6}{4} = \boxed{9} \end{aligned}$

$(\sin x + \mathrm{cosec} x)^2=\sin^2x+\mathrm{cosec}^2x+2 \\ +(\cos x + \sec x)^2=\cos^2x+\sec^2x+2 \\ \implies (\sin^2x+\cos^2x)+4+(\mathrm{cosec}^2x+\sec^2x) \\ 5+(\cot^2x+1+\tan^2x+1) \\ 7+\tan^2x+\cot^2x$

Now, the minimum value of $\tan^2x+\cot^2x$ is $2$ which can easily be shown by AM-GM inequality . $\therefore 7+2 = \large\boxed{9}$