Inspired from AIME problem 6

Lets first make everything into radians,

$s=\prod _{ k=1 }^{ 45 }{ ({ csc }^{ 2 } } (2k-1)\frac { \pi }{ 180 } \quad +3)$

$s=\prod _{ k=1 }^{ 45 }{ ({ cot }^{ 2 } } (2k-1)\frac { \pi }{ 180 } \quad +4)$

If we find a polynomial p(x) whose roots are ${ cot }^{ 2 }((2k-1)\frac { \pi }{ 180 } )$ all we have to do then is find p(-4).

The expression for tan(nx) is as follows

$tan(nx)=\frac { \left( \begin{matrix} n \\ 1 \end{matrix} \right) tan(x)-\left( \begin{matrix} n \\ 3 \end{matrix} \right) { tan }^{ 3 }(x)+.. }{ 1-\left( \begin{matrix} n \\ 2 \end{matrix} \right) { tan }^{ 2 }(x)+\left( \begin{matrix} n \\ 4 \end{matrix} \right) { tan }^{ 4 }(x)-.. }$

This expression can be obtained quite easily using complex no.s

We observe that for all $\theta =(2k-1)\pi /180$
$tan(90 \theta)$ = $tan((2k-1)\pi /2)$

which is undefined

This implies that for these values of $\theta$ the denominator for the expression of $tan(90 \theta)=0$

So we've got a polynomial in $tan(\theta)$ whose roots are $tan((2k-1)\pi /180)$ ,

$p(tan\quad \theta )=\left( \begin{matrix} 90 \\ 0 \end{matrix} \right) { tan }^{ 0 }(\theta )-\left( \begin{matrix} 90 \\ 2 \end{matrix} \right) { tan }^{ 2 }(\theta )+\left( \begin{matrix} 90 \\ 4 \end{matrix} \right) { tan }^{ 4 }(\theta )+...$

or $p(cot\quad \theta )=\left( \begin{matrix} 90 \\ 0 \end{matrix} \right) { cot }^{ 90 }(\theta )-\left( \begin{matrix} 90 \\ 2 \end{matrix} \right) cot^{ 88 }(\theta )+\left( \begin{matrix} 90 \\ 4 \end{matrix} \right) { cot }^{ 86 }(\theta )+..$

or $p({ cot }^{ 2 }\quad \theta )=\left( \begin{matrix} 90 \\ 0 \end{matrix} \right) { { cot }^{ 2 }(\theta ) }^{ 45 }-\left( \begin{matrix} 90 \\ 2 \end{matrix} \right) { { cot }^{ 2 }(\theta ) }^{ 44 }+\left( \begin{matrix} 90 \\ 4 \end{matrix} \right) { { cot }^{ 2 }(\theta ) }^{ 43 }+..$

Putting x= $cot^2(\theta)$ ,we get $p(x)={ x }^{ 45 }-\left( \begin{matrix} 90 \\ 2 \end{matrix} \right) { x }^{ 44 }+\left( \begin{matrix} 90 \\ 4 \end{matrix} \right) { x }^{ 43 }$ $=\prod _{ k=1\\ }^{ 45 }{ (x-{ cot }^{ 2 }( } (2k-1)\pi /180))$

Putting $x= -4$ ,we get

$s$ = $\sum _{ r=1 }^{ 45 }{ \left( \begin{matrix} 90 \\ 2r \end{matrix} \right) } { 2 }^{ 2r }$

= $\frac { { 3 }^{ 90 }+1 }{ 2 }$

Hence $2s-1={ 3 }^{ 90 }={ 9 }^{ 45 }$

$m=9,n=5$