4. If $a=\sin \theta+\tan \theta, b=\cos \theta+\cot \theta$, then which of the following expressions is incorrect? ( ).
(A) $\sin \theta=\frac{a b-1}{b+1}$
(B) $\cos \theta=\frac{1-a b}{a+1}$
(C) $\tan \theta+\cot \theta=\frac{(a+b+1)^{2}+1-2 a b}{(a+1)(b+1)}$
(D) $\tan \theta-\cot \theta=\frac{(a-b)(a+b+2)}{(a+1)(b+1)}$
4.B.

From the given, we have
$$
\begin{array}{l}
\sin \theta \cdot \cos \theta = a \cos \theta - \sin \theta, \\
\sin \theta \cdot \cos \theta = b \sin \theta - \cos \theta,
\end{array}
$$

Thus, \( a \cos \theta - \sin \theta = b \sin \theta - \cos \theta \).
Therefore, \((b+1) \sin \theta = (a+1) \cos \theta\), which gives
\(\tan \theta \doteq \frac{a+1}{b+1}\), \(\cot \theta = \frac{b+1}{a+1}\).
Hence, \(\tan \theta + \cot \theta = \frac{a+1}{b+1} + \frac{b+1}{a+1}\)
$$
\begin{array}{l}
= \frac{(a+b+1)^{2} + 1 - 2ab}{(a+1)(b+1)}, \\
\tan \theta - \cot \theta = \frac{a+1}{b+1} - \frac{b+1}{a+1} \\
= \frac{(a-b)(a+b+2)}{(a+1)(b+1)}.
\end{array}
$$

Therefore, options (C) and (D) are correct.
Substituting \(\tan \theta = \frac{a+1}{b+1}\) into the condition \(\sin \theta = a - \tan \theta = a - \frac{a+1}{b+1} = \frac{ab-1}{b+1}\), we know that option (A) is correct.

Substituting \(\cot \theta = \frac{b+1}{a+1}\) into the condition \(\cos \theta = b - \cot \theta = \frac{ab-1}{a+1}\), comparing with the value in option (B) will lead to \(ab-1=0\), making \(\cos \theta = 0\), at which point \(\tan \theta\) is undefined.