Step 1: Given $$\sin\theta + \cos\theta = \sqrt{3}$$
Step 2: Square both sides.
$$(\sin\theta + \cos\theta)^2 = (\sqrt{3})^2$$
$$\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 3$$
Step 3: Use the identity $\sin^2\theta + \cos^2\theta = 1$.
$$1 + 2\sin\theta\cos\theta = 3 \Rightarrow 2\sin\theta\cos\theta = 2$$
$$\Rightarrow \sin\theta\cos\theta = 1$$
Step 4: Now consider the LHS to be proved.
$$\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}$$
Step 5: Take LCM.
$$= \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}$$
Step 6: Substitute $\sin\theta\cos\theta = 1$ from Step 3.
$$\tan\theta + \cot\theta = \frac{1}{1} = 1$$
Hence proved.