Prove: `(tan^(3)theta-1)/(tan theta-1)=sec^(2)theta+tan theta`.

Explanation

To prove the identity \(\frac{\tan^3 \theta - 1}{\tan \theta - 1} = \sec^2 \theta + \tan \theta\), we can follow these steps:

### Step 1: Factor the left-hand side
We start with the expression \(\tan^3 \theta - 1\). We can use the identity for the difference of cubes:
\[ A^3 - B^3 = (A - B)(A^2 + AB + B^2) \]
Here, let \(A = \tan \theta\) and \(B = 1\). Thus, we have:
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