$\mathrm{A}=\tan {15}^0+\cot {15}^0, \mathrm{B}=\tan 22\frac{1^0}{2}+\cot 22\frac{1^0}{2}$ and $\mathrm{C} = \sin {54}^0-\sin {18}^0$ then the ascending order of A, B, C is

$\begin{array}{rcl}\mathrm{A}& =& \tan {15}^0+\cot {15}^0\\ & =& 2\mathrm{cosec}2\left({15}^0\right)\\ & =& 2\mathrm{cosec} {30}^0\\ & =& 2\left(2\right)=4\end{array}$

$\begin{array}{rcl}\mathrm{B}& =& \tan 22{\frac{1}{2}}^0+\cot 22{\frac{1}{2}}^0\\ & =& 2\mathrm{cosec}2\left(22{\frac{1}{2}}^0\right)\\ & =& 2\mathrm{cosec}{45}^0\\ & =& 2\sqrt{2}\end{array}$

$\begin{array}{rcl}\mathrm{C}& =& \sin {54}^0-\sin {18}^0\\ & =& \frac{\sqrt{5}+1}{4}-\frac{\sqrt{5}-2}{4}\\ & =& \frac{2}{4}=\frac{1}{2}=0.5\\ \mathrm{C}& <& \mathrm{B}<\mathrm{A}\end{array}$