Step 1: Divide the given equation by cos $\theta$ to bring in tan $\theta$ and sec $\theta$, instead of squaring it directly.
We are given:
\[ \sin\theta + \cos\theta = \sqrt{3} \]
Divide both sides by $\cos\theta$ (assuming $\cos\theta \neq 0$):
\[ \tan\theta + 1 = \sqrt{3}\sec\theta \]
Step 2: Square both sides to bring in the Pythagorean identity for sec and tan.
\[ (\tan\theta + 1)^2 = 3\sec^2\theta \]
\[ \tan^2\theta + 2\tan\theta + 1 = 3\sec^2\theta \]
Step 3: Replace sec squared theta using the identity 1 + tan squared theta = sec squared theta.
\[ \tan^2\theta + 2\tan\theta + 1 = 3(1 + \tan^2\theta) \]
\[ \tan^2\theta + 2\tan\theta + 1 = 3 + 3\tan^2\theta \]
Step 4: Bring all terms to one side to form a quadratic purely in tan $\theta$.
\[ 0 = 3 + 3\tan^2\theta - \tan^2\theta - 2\tan\theta - 1 \]
\[ 0 = 2\tan^2\theta - 2\tan\theta + 2 \]
Divide throughout by 2:
\[ \tan^2\theta - \tan\theta + 1 = 0 \]
Step 5: Rearrange this relation into the exact shape we need for tan $\theta$ + cot $\theta$.
From the equation above:
\[ \tan^2\theta + 1 = \tan\theta \quad \text{(Key Relation)} \]
Step 6: Write tan $\theta$ + cot $\theta$ as a single fraction and substitute the key relation.
\[ \tan\theta + \cot\theta = \tan\theta + \frac{1}{\tan\theta} = \frac{\tan^2\theta + 1}{\tan\theta} \]
Using the Key Relation from Step 5, the numerator $\tan^2\theta + 1$ is exactly equal to $\tan\theta$:
\[ \tan\theta + \cot\theta = \frac{\tan\theta}{\tan\theta} = 1 \]
Final Answer:
Hence, $\tan\theta + \cot\theta = 1$ is proved.
\[ \boxed{\tan\theta + \cot\theta = 1} \]