use the unit circle to evaluate the product (tan x + sec x)(tan x - sec x) when x = π/4. 1 1 + √2 1 - √2 3 -1

Explanation:

Step1: Expand the product

Using the difference - of - squares formula $(a + b)(a - b)=a^{2}-b^{2}$, we have $(\tan x+\sec x)(\tan x - \sec x)=\tan^{2}x-\sec^{2}x$.

Step2: Recall the trigonometric identity

We know the identity $1+\tan^{2}x=\sec^{2}x$, so $\tan^{2}x-\sec^{2}x=- 1$.

Step3: Evaluate at $x = \frac{\pi}{4}$

Since $\tan^{2}x-\sec^{2}x=-1$ for all $x$ where the functions are defined, when $x = \frac{\pi}{4}$, the value of $(\tan x+\sec x)(\tan x - \sec x)$ is also $-1$.

Answer:

$-1$