setting the value of $z=\cos\theta+i\sin\theta$, $$z^n+\frac{1}{z^n}$$$$=z^n+z^{-n}$$
$$=(\cos\theta+i\sin\theta)^n+(\cos\theta+i\sin\theta)^{-n}$$
Using d-Moivre's theorem,
$$=\cos(n\theta)+i\sin(n\theta)+\cos(-n\theta)+i\sin(-n\theta)$$
$$=\cos(n\theta)+i\sin(n\theta)+\cos(n\theta)-i\sin(n\theta)$$
$$=2\cos(n\theta)$$
Since $\sin^2x_{ij}$ is an arbitrary number in $[0,1]$, and $\cos^2x_{ij}=1-\sin^2x_{ij}$, let's rewrite your expression, with $p_{ij}\in[0,1]$:
$$\prod_{j=1}^n(1-\prod_{i=1}^m p_{ij})+\prod_{i=1}^m(1-\prod_{j=1}^n(1-p_{ij}))$$
Now, consider a matrix of independent events $E_{ij}$ on a given probability space, where the probability of $E_{ij}$ occurring is $p_{ij}$.
The first product is the probability of the event $A$: "in every column $j$, at least one event $E_{ij}$ does not occur".
The second product is the probability of the event $B$: "in every row $i$, at least one event $E_{ij}$ is occurring".
Now, we have $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
If $A$ is false, it means that there is a column where all events occur, but then $B$ must be true. Therefore, $P(A\cup B)=1$. And finally, we may write:
$$P(A)+P(B)-1=P(A)+P(B)-P(A\cup B)=P(A\cap B)\in[0,1]$$
Hence $P(A)+P(B)\ge1$, which is exactly your inequality.
Best Answer
If $\cos(2m+1)x=0,$
$(2m+1)x=(2n+1)\dfrac\pi2$ where $n$ is any integer
$x=\dfrac{(2n+1)\pi}{2(2m+1)}$ where $0\le n\le2m$
Now using this
$\cos(2m+1)x=2^{2m}\cos^{2m+1}x+\cdots+(-1)^m(2m+1)\cos x$
So, the roots of $$2^{2m}c^{2m+1}+\cdots+(-1)^m(2m+1)c=0$$ are $\cos\dfrac{2(2n+1)\pi}{(2m+1)},0\le n\le2m$
If $\cos x=0,n=m$
So, the roots of $$2^{2m}c^{2m}+\cdots+(-1)^m(2m+1)=0$$ are $\cos\dfrac{(2n+1)\pi}{2(2m+1)},0\le n\le2m, m\ne n$
$\implies2^{2m}\prod_{n=0,n\ne m}^{2m}\cos\dfrac{(2n+1)\pi}{2(2m+1)}=(-1)^m(2m+1)$
As $\pi-\dfrac{(2n+1)\pi}{2(2m+1)}=\dfrac{\pi(1+2(2m-n))}{2(2m+1)}$
$\implies2^{2m}\prod_{n=0}^{m-1}(-1)^m\cos^2\dfrac{(2n+1)\pi}{2(2m+1)}=(-1)^m(2m+1)$
Can you take it from here?