If in a triangle ABC, $\sin A:\sin C=\sin (A-B):\sin (B-C)$ then, $a^2:b^2:c^2$ are in 

Solution:

we have, 

$\begin{array}{l}\tan 82{\frac{1}{2}}^{\circ }=\cot 7\frac{1}{2^{\circ }}=\frac{\cos 7\frac{1}{2}}{\sin 7{\frac{1}{2}}^{\circ }}\\ =\frac{2{\cos }^{2}7\frac{1^{\circ }}{2}}{2\sin 7\frac{1^{\circ }}{2}\cos 7\frac{1}{2}}=\frac{1+\cos {15}^{\circ }}{\sin {15}^{\circ }}\\ =\frac{1+\frac{\sqrt{3}+1}{2\sqrt{2}}}{\sqrt{3}-1}\\ =\frac{2\sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1}\\ =\frac{(2\sqrt{2}+\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\\ =\frac{2\sqrt{6}+2\sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{2}\\ =\sqrt{6}+\sqrt{2}+\sqrt{4}+\sqrt{3}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}\end{array}$