Show that $ \tan (A + \theta) $ can be simplified to $- \cot \theta$ as A tends to $\frac{\pi}{2}$ So far I have used the identity, 
$$\tan\left(\frac{\pi}{2} + \theta\right) = \frac{\tan A + \tan \theta}  {1 - \tan A \tan \theta}$$
As $A  \to \frac{\pi}{2}$, $\tan A \to \infty$, so my reasoning is, $\infty + \tan \theta = \infty$, which gives:
$$\frac{\infty}  {- \infty  \tan \theta}$$
$\infty$ cancels to $-1$ leaving $\frac{-1}{\tan \theta}$ as the answer, or $-\cot \theta$
My approach seems a bit iffy, and I was hoping someone could confirm if this is right or not.
Thanks 

Using: 
$$\lim_{A \to \pi/2} \frac{- \tan A \tan \theta}{1 - \tan A \tan \theta} = 1:$$
$$\lim_{A \to \pi /2} \frac{\tan A + \tan \theta}{1 - \tan A \tan \theta} = \lim_{A \to \pi/2} \frac{\tan A}{1 - \tan A \tan \theta} + \frac{\tan \theta}{1 - \tan A \tan \theta} = \lim_{A \to \pi /2} \frac{\tan A}{-\tan A \tan \theta} + \frac{\tan \theta}{-\tan A \tan \theta} = \lim_{A \to \pi/2} \frac{-1}{\tan \theta} + \frac{-1}{\tan A} = -\frac{1}{\tan \theta} = -\cot \theta$$
