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 
 A: $\displaystyle \lim_{A \to \dfrac{\pi}{2}} \tan(A+\theta)=\displaystyle \lim_{A\to \dfrac{\pi}{2}} \dfrac{\sin(A+\theta)}{\cos (A+\theta)}=\dfrac{\sin(\dfrac{\pi}{2}+\theta)}{\cos(\dfrac{\pi}{2}+\theta)}=\dfrac{\cos \theta}{-\sin \theta}=-\cot \theta$
A: Notice, we have $$\lim_{A\to \pi/2}\tan(A+\theta)$$ $$=\lim_{A\to \pi/2}\frac{\tan A+\tan\theta}{1-\tan A\tan \theta}$$ $$=\lim_{A\to \pi/2}\frac{1+\frac{\tan\theta}{\tan A}}{\frac{1}{\tan A}-\tan \theta}$$ Let, $\frac{1}{\tan A}=t\implies t\to 0\ as \ A\to \frac{\pi}{2}$$$=\lim_{t\to 0}\frac{1+(t)\tan\theta}{t-\tan \theta}$$   $$=\frac{1+0}{0-\tan \theta}$$ $$=-\frac{1}{\tan \theta}=-\cot \theta$$
A: Using: 
$$\lim_{A \to \pi/2} \frac{- \tan A \tan \theta}{1 - \tan A \tan \theta} = 1:$$
$$\lim_{A \to \pi /2} \frac{\tan A + \tan \theta}{1 - \tan A \tan \theta} = \lim_{A \to \pi/2} \frac{\tan A}{1 - \tan A \tan \theta} + \frac{\tan \theta}{1 - \tan A \tan \theta} = \lim_{A \to \pi /2} \frac{\tan A}{-\tan A \tan \theta} + \frac{\tan \theta}{-\tan A \tan \theta} = \lim_{A \to \pi/2} \frac{-1}{\tan \theta} + \frac{-1}{\tan A} = -\frac{1}{\tan \theta} = -\cot \theta$$
A: As I mentioned in a comment, the way to (try to) make your approach rigorous is to divide numerator and denominator both by the value which goes to infinity, namely $\tan(A)$, leaving limits which can be evaluated without such shenanigans:
$$\frac{\tan(A)+\tan(\theta)}{1-\tan(A)\tan(\theta)}
=\frac{1+\cot(A)\tan(\theta)}{\cot(A)-\tan(\theta)}$$
Now the limit as $A\to\frac\pi2$ can be evaluated directly without any terms blowing up.
