Applying every concept we know

We start by considering the complex polynomial

$p(z)=z^{n}-e^{\iota\theta}$

The roots of this polynomial are $e^{\iota(\theta+2k\pi)/n}, 0\leq k\leq n-1$ and $k\in\mathbb{Z}$ .

Therefore the polynomial can be written as

$p(z)=(z-e^{\iota\theta/n})(z-e^{\iota(\theta+2\pi)/n})(z-e^{\iota(\theta+4\pi)/n})...(z-e^{\iota(\theta+2(n-1)\pi)/n})$

Substituting $z=1$ we get

$1-e^{\iota\theta}=(1-e^{\iota\theta/n})\displaystyle\prod_{k=1}^{n-1} \left( -2\iota\sin\left( \dfrac{\theta+2k\pi}{2n}\right)e^{\iota(\theta+2k\pi)/2n}\right)$

On simplifying the above identity we get

$\displaystyle\prod_{k=1}^{n-1} \sin\left( \dfrac{\theta+2k\pi}{2n}\right)=\dfrac{\sin (\theta/2)}{2^{n-1}\sin (\theta/2n)}$

$\therefore\displaystyle\sum_{k=1}^{n-1} \ln\left( \sin\left( \dfrac{\theta+2k\pi}{2n}\right)\right)=\ln\sin\dfrac{\theta}{2}-\ln\sin\dfrac{\theta}{2n}+\ln\left( \dfrac{1}{2^{n-1}}\right)$

Differentiating both sides w.r.t. $\theta$

$\displaystyle\sum_{k=1}^{n-1}\cot\left( \dfrac{\theta+2k\pi}{2n}\right)=n\cot\dfrac{\theta}{2}-\cot\dfrac{\theta}{2n}$

Again differentiating w.r.t. $\theta$

$\displaystyle\sum_{k=1}^{n-1}\csc^{2}\left( \dfrac{\theta+2k\pi}{2n}\right)=n^{2}\csc^{2}\dfrac{\theta}{2}-\csc^{2}\dfrac{\theta}{2n}$

Using the identity $1+\cot^{2}\theta=\csc^{2}\theta$ we get

$\displaystyle\sum_{k=1}^{n-1}\cot^{2}\left( \dfrac{\theta+2k\pi}{2n}\right)=n^{2}\csc^{2}\dfrac{\theta}{2}-\csc^{2}\dfrac{\theta}{2n}-n+1$

Now substitute $\theta=\pi,n=50$ to obtain

$\displaystyle\sum_{k=1}^{49}\cot^{2}\left( \dfrac{(2k+1)\pi}{100}\right)=2451-\csc^{2}\dfrac{\pi}{100}$

$\therefore a+b=\boxed{2551}$