Establish the identity. $$ (\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1 $$ Multiply and write the left side expression as the difference of two squares.

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To establish the identity

$$
(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1,
$$

we can use the **difference of squares** formula, which states that:

$$
(a + b)(a - b) = a^2 - b^2.
$$

Applying this to the left side of the identity:

$$
(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = (\sec \theta)^2 - (\tan \theta)^2.
$$

Now, recall the Pythagorean identity for secant and tangent:

$$
\sec^2 \theta - \tan^2 \theta = 1.
$$

Substituting this into our expression:

$$
(\sec \theta)^2 - (\tan \theta)^2 = 1.
$$

Therefore,

$$
(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1.
$$

This confirms that the original identity holds true.

**Answer:**  
After expanding, the left side is sec²θ – tan²θ, which equals 1. Thus,

(secθ + tanθ)(secθ – tanθ) = sec²θ − tan²θ = 1.
$$ (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = \sec^2 \theta - \tan^2 \theta = 1 $$

Establish the identity. \[ (\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1 \] Multiply and write the left side expression as the difference of two squares.

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