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