Looks like what?

Let's Evaluate $\color{#3D99F6}N$ First. Using Trigonometric Identity

$\tan{(A+B)}=\dfrac{\tan{A}+\tan{B}}{1-\tan{A}\times\tan{B}}$

$\tan{(40+20)}=\dfrac{\tan{40}+\tan{20}}{1-\tan{40}\tan{20}} \implies \tan{40}+\tan{20}=\tan{60}(1-\tan{40}\tan{20}) \\ \tan{40}+\tan{20}=\sqrt{3}-\sqrt{3}\tan{40}\tan{20} \\ \color{#3D99F6}N=\boxed{\sqrt{3}}=\tan{40}+\tan{20}+\sqrt{3}\tan{40}\tan{20}$

To Find the Value of $\color{#20A900}M$ . We will use the identity of $\cot{(A+B)}$ and $\cot{(A-B)}$

$\cot{(A\pm B)}=\dfrac{\cot{B}\times\cot{A}\mp1}{\cot{B}\pm\cot{A}} \\ \cot{(16+44)}=\dfrac{\cot{44}\cot{16}-1}{\cot{44}+\cot{16}} \\ \implies \cot{44}\cot{16}=\dfrac1{\sqrt3}(\cot{44}+\cot{16})+1\\ \cot{(44+76)}=\dfrac{\cot{76}\cot{44}-1}{\cot{76}+\cot{44}} \\ \implies \cot{76}\cot{44}= -\dfrac1{\sqrt3}(\cot{76}+\cot{44})+1\\ \cot{(76-16)}=\dfrac{\cot{16}\cot{76}+1}{\cot{16}-\cot{76}} \\ \implies \cot{16}\cot{76}=\dfrac1{\sqrt3}(\cot{16}-\cot{76})-1 \\ \implies -\cot{16}\cot{76}=\dfrac1{\sqrt3}(\cot{76}-\cot{16})+1 \\ \color{#20A900}M=\cot{16}\cot{44}+\cot{44}\cot{76}-\cot{76}\cot{16} \\ \implies \color{#20A900}M=\dfrac1{\sqrt3}(\cot{44}+\cot{16})+1+\dfrac1{\sqrt3}(\cot{76}-\cot{16})-\dfrac1{\sqrt3}(\cot{76}+\cot{44})+1 = \boxed{3} \\ \dfrac{\color{#20A900}M}{\color{#3D99F6}N^2}=\boxed{1}$