Answer
$\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=2sec^{2}\theta$
Work Step by Step
Need to prove the identity $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=2sec^{2}\theta$
Let us take left side solve first.
$\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=\frac{1+sin\theta}{(1-sin\theta)({1+sin\theta)}}+\frac{1-sin\theta}{(1-sin\theta)({1+sin\theta)}}$
$\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=\frac{2}{1-sin^{2}\theta}$
Since, $sin^{2}\theta+cos^{2}\theta=1$
$\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=\frac{2}{cos^{2}\theta}$
But $\frac{1}{cos\theta}=sec\theta$
Hence, $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}=2sec^{2}\theta$
