A geometry problem by Joyce Mao

$\begin{aligned} f(a) & = \frac {\sin^3 a+ 6\sin^2 a + \sin a + 2{\color{#3D99F6}\cos^2 a}-8}{\sin a -1} \\ & = \frac {\sin^3 a+ 6\sin^2 a + \sin a + 2{\color{#3D99F6}(1- \sin^2 a)}-8}{\sin a -1} \\ & = \frac {\sin^3 a+ 4\sin^2 a + \sin a -6}{\sin a -1} & \small \color{#3D99F6} \text{By long division} \\ & = \sin^2 a + 5 \sin a + 6 \\ & = {\color{#3D99F6}\left(\sin a + \frac 52\right)^2} - \frac {25}4 + 6 & \small \color{#3D99F6} \text{Since }\max \left( \left({\color{#D61F06}\sin a} + \frac 52\right)^2 \right) = \left({ \color{#D61F06}1} + \frac 52\right)^2 \\ & = {\color{#3D99F6}\frac {49}4} - \frac {25}4 + 6 \\ & = 12 \end{aligned}$

But when $\sin a = 1$ , $f(a)$ is undefined. Therefore, the maximum can't be determined .