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.

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.