Answer
$$\frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}=2\sec^2\theta$$
The left side has been proved to be equal with the right side. It is thus an identity.
Work Step by Step
$$\frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}=2\sec^2\theta$$
The left side looks more complex, so we deal with the left side first.
$$A=\frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}$$
$$A=\frac{1+\sin\theta+1-\sin\theta}{(1-\sin\theta)(1+\sin\theta)}$$
$$A=\frac{2}{(1-\sin\theta)(1+\sin\theta)}$$
$$A=\frac{2}{1-\sin^2\theta}$$
(for $(a-b)(a+b)=a^2-b^2$)
Now we can solve $1-\sin^2\theta$ by using a Pythagorean Identity:
$$\cos^2\theta=1-\sin^2\theta$$
$$A=\frac{2}{\cos^2\theta}$$
From a Reciprocal Identity:
$$\sec\theta=\frac{1}{\cos\theta}$$
So, $$\sec^2\theta=\frac{1}{\cos^2\theta}$$
Therefore, $$A=2\sec^2\theta$$
The trigonometric expression is hence an identity.
