Trigonometry (11th Edition) Clone

$$\sec^2\theta-1=(\sec\theta-1)(\sec\theta+1)$$
$$A=\sec^2\theta-1$$ We factor the expression like normal, using $$a^2-b^2=(a-b)(a+b)$$ Therefore, $$A=(\sec\theta-1)(\sec\theta+1)$$