Answer
$$(\sec\alpha-\tan\alpha)^2=\frac{1-\sin\alpha}{1+\sin\alpha}$$
It is an identity, by representing the left side in terms of $\sin\alpha$.
Work Step by Step
$$(\sec\alpha-\tan\alpha)^2=\frac{1-\sin\alpha}{1+\sin\alpha}$$
In this exercise, we would deal with the left side first.
$$A=(\sec\alpha-\tan\alpha)^2$$
- Reciprocal Identity: $$\sec\alpha=\frac{1}{\cos\alpha}$$
- Quotient Identity: $$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$$
Therefore, $$\sec\alpha-\tan\alpha=\frac{1}{\cos\alpha}-\frac{\sin\alpha}{\cos\alpha}=\frac{1-\sin\alpha}{\cos\alpha}$$
$A$ would thus be $$A=\Big(\frac{1-\sin\alpha}{\cos\alpha}\Big)^2$$
$$A=\frac{(1-\sin\alpha)^2}{\cos^2\alpha}$$
Using Pythagorean Identity for $\cos^2\alpha$:
$$\cos^2\alpha=1-\sin^2\alpha$$
$$A=\frac{(1-\sin\alpha)^2}{1-\sin^2\alpha}$$
$$A=\frac{(1-\sin\alpha)^2}{(1-\sin\alpha)(1+\sin\alpha)}$$ (for $a^2-b^2=(a-b)(a+b)$)
$$A=\frac{1-\sin\alpha}{1+\sin\alpha}$$
The right side has been proved to be similar with the left side. The trigonometric expression is an identity.
