Answer: B
\(\sin^2 \theta + \csc^2 \theta\) \(= (\sin \theta + \csc \theta)^2 - 2\sin\theta \csc\theta\) \(= 2^2 - 2 \times \cancel{\sin\theta} \times \frac{1}{\cancel{\sin\theta}}\) \(= 4 - 2 = 2\)
\(\sin^2 \theta + \csc^2 \theta\) \(= (\sin \theta + \csc \theta)^2 - 2\sin\theta \csc\theta\) \(= 2^2 - 2 \times \cancel{\sin\theta} \times \frac{1}{\cancel{\sin\theta}}\) \(= 4 - 2 = 2\)