Cotan Product... Sort of...

$\Large{\displaystyle\prod_{r=1}^{44} (\cot r - 1) \\ = \displaystyle \prod_{r=1}^{44} \left(\dfrac{\cos r}{\sin r} - 1\right) \\ = \displaystyle\prod_{r=1}^{44} \left(\dfrac{\cos r - \sin r}{\sin r}\right) \\ = \displaystyle\prod_{r=1}^{44} \dfrac{\sin(90-r) - \sin r}{\sin r} \\ = \displaystyle\prod_{r=1}^{44} \dfrac{2\cos\left(\dfrac{90}{2}\right)\sin\left(\dfrac{90-2r}{2}\right)}{\sin r} \\ = \displaystyle\prod_{r=1}^{44} \dfrac{\sqrt{2}\sin(45-r)}{\sin r} \\ = (\sqrt{2})^{44} \times \dfrac{\sin 44^\circ\times \sin 43^\circ\times \dots \sin 23^\circ\times \sin 22^\circ\times \dots \sin 2^\circ\times\sin 1^\circ}{\sin 1^\circ\times\sin 2^\circ\times \dots \sin 22^\circ\times\sin 23^\circ\times \dots\sin 43^\circ\times\sin 44^\circ} \\ \\ \\ = 2^{\frac{1}{2} \times 44} \times 1 \\ = 2^{22} \\ \Rightarrow \boxed{k=22}}$

Note: All the angles mentioned are in degrees

$\large{1) \quad \tan (45-x)=\frac{1-\tan (x)}{1+\tan (x)}}$

$\large{2) \quad \prod _{ r=1 }^{ 44 }{ \frac { \tan { \left( 45-r \right) } }{ \tan { \left( r \right) } } } =\frac{\color{#D61F06}{\tan(44)}\color{#3D99F6}{\tan(43)}.......\color{#20A900}{\tan(1)}}{\color{#20A900}{\tan(1)}.......\color{#3D99F6}{\tan(43)}\color{#D61F06}{\tan(44)}}=1}$

$\large{3) \quad (1+\tan(x))(1+\tan(45-x))=1+\tan(x)+\tan(45-x)+\tan(x)\tan(45-x)=1+1-\tan(x)\tan(45-x)+\tan(x)\tan(45-x)=2}$

Now to the question

$\large{\prod _{ r=1 }^{ 44 }{ \frac { 1-\tan { \left( r \right) } }{ \tan { \left( r \right) } } } }$

$\large{=\prod _{ r=1 }^{ 44 }{ \frac { \tan { \left( 45-r \right) \left( 1+\tan { \left( r \right) } \right) } }{ \tan { \left( r \right) } } } }$

$\large{=\prod _{ r=1 }^{ 44 }{ \frac { \tan { \left( 45-r \right) } }{ \tan { \left( r \right) } } } \prod _{ r=1 }^{ 44 }{ \left( 1+\tan { \left( r \right) } \right) } }$

$\large{=(1+\tan(1))(1+\tan(44))(1+\tan(2))(1+\tan(43)).......(1+\tan(22))(1+\tan(23))}$

we observe that there are $22$ pairs which follow statement $3$ above.

thus the answer is $\large{\boxed{2^{22}}}$