If $20 {\sin }^2 \mathrm{\alpha }+21\cos \mathrm{\alpha }-24=0$ where 

$\frac{7\mathrm{\pi }}{4}<\mathrm{\alpha }<2\mathrm{\pi }$ then $\cot \frac{\mathrm{\alpha }}{2}=$

$$\begin{gathered} 20 {\sin }^2 \mathrm{\alpha }+21\cos \mathrm{\alpha }-24=0 \\ 20-20{\cos }^2\mathrm{\alpha }+21\cos \mathrm{\alpha }-24=0 \\ 20{\cos }^2\mathrm{\alpha }-21\cos \mathrm{\alpha }+4=0 \\ \left(4\cos \mathrm{\alpha }-1\right) \left(5\cos \mathrm{\alpha }-4\right)=0 \\ \cos \mathrm{\alpha }=\frac{1}{4}, \cos \mathrm{\alpha }=\frac{4}{5} \\ \frac{1-{\tan }^2 {\displaystyle \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{1+{\tan }^2 {\displaystyle \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{1}{4} \\ 4-4{\tan }^2 \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=1+{\tan }^2 \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ 5{\tan }^2 \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=-3 \end{gathered}$$

which is not possible

$$\begin{gathered} \cos \mathrm{\alpha }=\frac{4}{5} \\ \frac{1-{\tan }^2 {\displaystyle \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{1+{\tan }^2 {\displaystyle \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{4}{5} \\ 5-5 {\tan }^2 \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=4+4{\tan }^2 \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ 9 {\tan }^2 \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=1 \\ {\cot }^2 \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=9 \Rightarrow \cot \raisebox{1ex}{$\mathrm{\alpha }$}\!\left/ \!\raisebox{-1ex}{$2$}\right. =-3 \end{gathered}$$