Answer
$$(\sec\theta-1)(\sec\theta+1)=\tan^2\theta$$
Work Step by Step
$$A=(\sec\theta-1)(\sec\theta+1)$$
As $(a-b)(a+b)=a^2-b^2$:
$$A=\sec^2\theta-1$$
- Reciprocal Identity: $$\sec\theta=\frac{1}{\cos\theta}$$
Replace into $A$:
$$A=\frac{1}{\cos^2\theta}-1$$
$$A=\frac{1-\cos^2\theta}{\cos^2\theta}$$
- Pythagorean Identity:
$$\sin^2\theta=1-\cos^2\theta$$
Replace into $A$:
$$A=\frac{\sin^2\theta}{\cos^2\theta}=\Bigg(\frac{\sin\theta}{\cos\theta}\Bigg)^2$$
- Quotient Identity: $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$
Replace into $A$:
$$A=\tan^2\theta$$
