Answer
$$\frac{1-\sin\theta}{1+\sin\theta}=\sec^2\theta-2\sec\theta\tan\theta+\tan^2\theta$$
The trigonometric expression is an identity.
Work Step by Step
$$\frac{1-\sin\theta}{1+\sin\theta}=\sec^2\theta-2\sec\theta\tan\theta+\tan^2\theta$$
We would deal with the right side first.
$$A=\sec^2\theta-2\sec\theta\tan\theta+\tan^2\theta$$
We can use $$a^2-2ab+b^2=(a-b)^2$$
That means $$A=(\sec\theta-\tan\theta)^2$$
Now we transform $\sec\theta$ and $\tan\theta$ according to the identities:
$$\sec\theta=\frac{1}{\cos\theta}\hspace{2cm}\tan\theta=\frac{\sin\theta}{\cos\theta}$$
Therefore $A$ would be
$$A=\Big(\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}\Big)^2$$
$$A=\Big(\frac{1-\sin\theta}{\cos\theta}\Big)^2$$
$$A=\frac{(1-\sin\theta)^2}{\cos^2\theta}$$
From Pythagorean Identity: $\cos^2\theta=1-\sin^2\theta=(1-\sin\theta)(1+\sin\theta)$
That makes $A$ into
$$A=\frac{(1-\sin\theta)^2}{(1-\sin\theta)(1+\sin\theta)}$$
$$A=\frac{1-\sin\theta}{1+\sin\theta}$$
The left side is therefore equal with the right side. The trigonometric expression is thus an identity.
