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

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To establish the trigonometric identity:

$$
(\csc \theta + \cot \theta)(\csc \theta - \cot \theta) = 1
$$

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

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

Let's apply this formula to the left side of the identity:

$$
(\csc \theta + \cot \theta)(\csc \theta - \cot \theta) = \csc^2 \theta - \cot^2 \theta
$$

Now, recall the fundamental Pythagorean identity in trigonometry:

$$
\csc^2 \theta = 1 + \cot^2 \theta
$$

Using this identity, we can substitute $$ \csc^2 \theta $$:

$$
\csc^2 \theta - \cot^2 \theta = (1 + \cot^2 \theta) - \cot^2 \theta = 1
$$

**Conclusion:**

$$
(\csc \theta + \cot \theta)(\csc \theta - \cot \theta) = \csc^2 \theta - \cot^2 \theta = 1
$$

This confirms that the original identity holds true.
$$ (\csc \theta + \cot \theta)(\csc \theta - \cot \theta) = 1 $$

Establish the identity.

Multiply and write the left side expression as the difference of two squares.

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Establish the identity. Multiply and write the left side expression as the difference of two squares.