2012-II-19

Ans: C
$\begin{array}{cl}
& \dfrac{\cos60^\circ}{1-\cos(90^\circ-\theta)}+\dfrac{\cos240^\circ}{1-\cos(270^\circ-\theta)} \\
= & \dfrac{\cos60^\circ}{1-\sin\theta}+\dfrac{-\cos60^\circ}{1+\sin\theta} \\
= & \dfrac{\cos60^\circ(1+\sin\theta)-\cos60^\circ(1-\sin\theta)}{(1-\sin\theta)(1+\sin\theta)} \\
= & \dfrac{2\cos60^\circ\sin\theta}{1-\sin^2 \theta} \\
= & \dfrac{\sin\theta}{\cos^2\theta} \\
= & \dfrac{\sin\theta}{\cos\theta\times\cos\theta} \\
= & \dfrac{\tan\theta}{\cos\theta}
\end{array}$