Ans: A
$\begin{array}{cl}
& \dfrac{\sin\theta}{\cos 60^\circ} +\dfrac{\cos (270^\circ-\theta)}{\tan 45^\circ} \\
= & \dfrac{\sin\theta}{\frac{1}{2}}+\dfrac{-\sin\theta}{1} \\
= & 2\sin\theta-\sin\theta \\
= & \sin\theta
\end{array}$
$\begin{array}{cl}
& \dfrac{\sin\theta}{\cos 60^\circ} +\dfrac{\cos (270^\circ-\theta)}{\tan 45^\circ} \\
= & \dfrac{\sin\theta}{\frac{1}{2}}+\dfrac{-\sin\theta}{1} \\
= & 2\sin\theta-\sin\theta \\
= & \sin\theta
\end{array}$
