Step 1: Rewrite the expression we need to prove, in terms of sine and cosine, before touching the given equation.
We are given $\sin\theta + \cos\theta = \sqrt{3}$ and asked to prove $\tan\theta + \cot\theta = 1$. Instead of squaring the given equation first, start by simplifying the left side of what we want to prove.
Write $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\cot\theta = \frac{\cos\theta}{\sin\theta}$, so:
\[ \tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} \]
Combine these two fractions over the common denominator $\sin\theta\cos\theta$:
\[ \tan\theta + \cot\theta = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} \]
Using the identity $\sin^2\theta + \cos^2\theta = 1$, this becomes a general fact true for any suitable angle $\theta$:
\[ \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta} \quad \text{(Key identity)} \]
So proving $\tan\theta + \cot\theta = 1$ is the same as proving $\sin\theta\cos\theta = 1$.
Step 2: Now bring in the given equation to find the needed product.
We are given:
\[ \sin\theta + \cos\theta = \sqrt{3} \]
Square both sides:
\[ (\sin\theta + \cos\theta)^2 = (\sqrt{3})^2 \]
\[ \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 3 \]
Substitute $\sin^2\theta + \cos^2\theta = 1$:
\[ 1 + 2\sin\theta\cos\theta = 3 \]
\[ 2\sin\theta\cos\theta = 2 \]
\[ \sin\theta\cos\theta = 1 \]
Step 3: Substitute this value into the key identity from Step 1.
From Step 1:
\[ \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta} \]
Since $\sin\theta\cos\theta = 1$:
\[ \tan\theta + \cot\theta = \frac{1}{1} = 1 \]
This is exactly the result we needed to prove.
Final Answer:
By first turning $\tan\theta + \cot\theta$ into $\frac{1}{\sin\theta\cos\theta}$ and then evaluating $\sin\theta\cos\theta$ from the given condition, we get $\tan\theta + \cot\theta = 1$, proving the required result.