Answer
\[n_{1}\sin\theta_{1} = n_{2}\sin\theta_{2}\]
Work Step by Step
\[\frac{n_{1}}{\sqrt{\cot^2 \theta_{1}+1}} = \frac{n_{2}}{\sqrt{\cot^2 \theta_{2}+1}}\]
where we have that $\cot\theta_{1} = \frac{\cos\theta_{1}}{\sin\theta_{1}}$ and $\cot\theta_{2} = \frac{\cos\theta_{2}}{\sin\theta_{2}}$.
\[\implies \frac{n_{1}}{\sqrt{\frac{\cos^2\theta_{1}}{\sin^2\theta_{1}}+1}} = \frac{n_{2}}{\sqrt{\frac{\cos^2\theta_{2}}{\sin^2\theta_{2}}+1}}\]
\[\implies \frac{n_{1}}{\sqrt{\frac{\cos^2\theta_{1}+\sin^2\theta_{1}}{\sin^2\theta_{1}}}} = \frac{n_{1}}{\sqrt{\frac{\cos^2\theta_{2}+\sin^2\theta_{2}}{\sin^2\theta_{2}}}}\]
where $\cos^2\theta_{1}+\sin^2\theta_{1} = 1$ and $\cos^2\theta_{2}+\sin^2\theta_{2} = 1$.
\[\implies n_{1}\sin\theta_{1} = n_{2}\sin\theta_{2}\]
