Solving the Trigonometric Equation

We are given the equation $1 + \tan \theta = \sqrt{3}$. Our goal is to find the value of the expression $\sqrt{3} \cot \theta - 1$.

Step 1: Determine $\tan \theta$

Rearrange the initial equation to solve for $\tan \theta$:

$1 + \tan \theta = \sqrt{3}$

Subtract 1 from both sides:

$\tan \theta = \sqrt{3} - 1$

Step 2: Calculate $\cot \theta$

Use the identity $\cot \theta = \frac{1}{\tan \theta}$:

$\cot \theta = \frac{1}{\sqrt{3} - 1}

Rationalize the denominator by multiplying by the conjugate $(\sqrt{3} + 1)$:

$\cot \theta = \frac{1}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} = \frac{\sqrt{3} + 1}{(\sqrt{3})^2 - 1^2} = \frac{\sqrt{3} + 1}{3 - 1} = \frac{\sqrt{3} + 1}{2}

Step 3: Evaluate $\sqrt{3} \cot \theta - 1$

Substitute the value of $\cot \theta$ into the expression:

$\sqrt{3} \cot \theta - 1 = \sqrt{3} \left( \frac{\sqrt{3} + 1}{2} \right) - 1$

Simplify the multiplication:

$= \frac{\sqrt{3}(\sqrt{3} + 1)}{2} - 1 = \frac{3 + \sqrt{3}}{2} - 1$

Combine the terms by using a common denominator:

$= \frac{3 + \sqrt{3}}{2} - \frac{2}{2} = \frac{3 + \sqrt{3} - 2}{2} = \frac{1 + \sqrt{3}}{2}

The final result is:

$\frac{\sqrt{3} + 1}{2}$