Statement-1: ${\mathrm{NO}}_2^{-}$ ion can be detected by brown ring test in presence of  ${\mathrm{NO}}_3^{-}$ion.

Statement-2: Both ${\mathrm{NO}}_2^{-}$ and ${\mathrm{NO}}_3^{-}$ion are confirmed by brown ring test.

The brown ring test for ${\mathrm{NO}}_2^{-}$ and ${\mathrm{NO}}_3^{-}$ is due to the formation of complex,  $\left[\mathrm{Fe}\right(\mathrm{H}2\mathrm{O}{)}_5{\mathrm{NO}}^2+$ The procedure used to detect the presence of nitrate radicals is known as the Brown ring test.

For this, an equivalent amount of freshly made ${\mathrm{FeSO}}_4$is combined with the sample solution. Along the test tube's sides, concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ is added to the mixture. The two liquids will mix, forming a brown ring.

This substance is nitro so Ferrous Sulphate $\left({\mathrm{FeSO}}_4\cdot \mathrm{NO}\right)$Ferrous is converted into ferric by nitric acid (nitrate), which is then converted into nitrogen monoxide (NO). The brown ring is the result of this reaction with ${\mathrm{FeSO}}_4$.

Hence, the correct option is B.