Here, $\mathrm{AB}=6 \mathrm{~cm}, \mathrm{BC}=8 \mathrm{~cm}$ and $\mathrm{CA}=10 \mathrm{~cm}$
So, $c=6 \mathrm{~cm}, \mathrm{a}=8 \mathrm{~cm}, \mathrm{~b}=10 \mathrm{~cm}$
$\mathrm{S}=\frac{a+b+c}{2}=\frac{24}{2}=12$
$\therefore \tan \frac{A}{2}=\sqrt{\frac{(12-10)(12-6)}{12(12-8)}}$
$\left(\because \tan \frac{\mathrm{A}}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\right)$
$=\sqrt{\frac{1}{4}}=\frac{1}{2}$


$\therefore \cot \frac{A}{2}=2$
Now, $\cot \left(\frac{A}{4}+\frac{A}{4}\right)=\frac{\cot ^{2} \frac{A}{4}-1}{2 \cot \frac{A}{4}}$
$\cot \left(\frac{A}{2}\right)=\frac{\cot ^{2} \frac{A}{4}-1}{2 \cot \frac{A}{4}}$
Let $\cot \left(\frac{A}{4}\right)=x$
$\therefore \quad 2=\frac{x^{2}-1}{2 x}$
$\Rightarrow x^{2}-4 x-1=0$
$\Rightarrow x=\frac{4 \pm \sqrt{16+4}}{2}$
$\Rightarrow x=\frac{4 \pm 2 \sqrt{5}}{2}=2 \pm \sqrt{5}$
So, $\cot \left(\frac{A}{4}\right)=\sqrt{5}+2$ or $2-\sqrt{5}$