Chapter 14 Trigonometric Graphs, Identities, and Equations - 14.3 Verify Trigonometric Identities - 14.3 Exercises - Problem Solving - Page 930: 45b

\[\frac{n^2_{2}-n^2_{1}}{n^2_{2}}\sin^2\theta_{1} = \frac{n^2_{2}-n^2_{1}}{n^2_{1}}\cos^2\theta_{1}\]

From (a), \[\frac{n^2_{1}}{n^2_{2}}\sin^2\theta_{1} +\frac{n^2_{2}}{n^2_{1}}\cos^2\theta_{1} = 1\] Let us rewrite the right hand side $1$ as $\sin^2\theta_{1}+\cos^2\theta_{1}$. \[\implies \frac{n^2_{1}}{n^2_{2}}\sin^2\theta_{1} +\frac{n^2_{2}}{n^2_{1}}\cos^2\theta_{1} = \sin^2\theta_{1}+\cos^2\theta_{1} \] Simplifying further \[\frac{n^2_{1}}{n^2_{2}}\sin^2\theta_{1}-\sin^2\theta_{1} = \cos^2\theta_{1}-\frac{n^2_{2}}{n^2_{1}}\cos^2\theta_{1}\] \[\implies \frac{n^2_{1}-n^2_{2}}{n^2_{2}}\sin^2\theta_{1} = \frac{n^2_{1}-n^2_{2}}{n^2_{1}}\cos^2\theta_{1}\] \[\implies \frac{n^2_{2}-n^2_{1}}{n^2_{2}}\sin^2\theta_{1} = \frac{n^2_{2}-n^2_{1}}{n^2_{1}}\cos^2\theta_{1}\] which is the required answer.