## Calculus: Early Transcendentals 8th Edition

$\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=2sec^{2}\theta$
We need to prove the identity $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=2sec^{2}\theta$ $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=\frac{1+sin\theta}{(1-sin\theta)({1+sin\theta)}}+\frac{1-sin\theta}{(1-sin\theta)({1+sin\theta)}}$ $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=\frac{2}{1-sin^{2}\theta}$ Since $sin^{2}\theta+cos^{2}\theta=1$: $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=\frac{2}{cos^{2}\theta}$ Also, $\frac{1}{cos\theta}=sec\theta$ Hence, $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=2sec^{2}\theta$