If $\sin \theta + \cos \theta = \sqrt{3}$,then prove that $\tan \theta + \cot \theta = 1$.

(N/A) Given: $\sin \theta + \cos \theta = \sqrt{3}$.
Squaring both sides:
$(\sin \theta + \cos \theta)^2 = (\sqrt{3})^2$
$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 3$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have:
$1 + 2 \sin \theta \cos \theta = 3$
$2 \sin \theta \cos \theta = 2$
$\sin \theta \cos \theta = 1$.
Now,consider the expression $\tan \theta + \cot \theta$:
$\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$.
Substituting $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin \theta \cos \theta = 1$:
$\tan \theta + \cot \theta = \frac{1}{1} = 1$.
Hence,proved.