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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

Michael Hardy
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Jack
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  • Iffy? Not at all. Cancelling out infinities perfectly makes sense. –  Jul 24 '15 at 01:05
  • You cannot cancel out the infinities like that. – Rick Jul 24 '15 at 01:12
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    @BolzWeir I vehemently disagree. After all, you could say that as $x\to\infty$, $\frac{x}{2x}\to\frac{\infty}{\infty}=1$. The way the notion can be made rigorous here is to divide numerator and denominator by $\tan(A)$so that none of the terms actually tend to infinite values. – Steven Stadnicki Jul 24 '15 at 01:13
  • @StevenStadnicki: I'm just being sarcastic. –  Jul 24 '15 at 01:16
  • @Jack, for future posts, you can have a read through http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference. – John_dydx Jul 24 '15 at 01:21
  • Thank you everyone for your great answers. I feel a bit silly with that 'infinity logic' which @StevenStadnicki cleared up. John I will definitely go over that before my next post! Thanks again. – Jack Jul 24 '15 at 01:37

4 Answers4

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$\displaystyle \lim_{A \to \dfrac{\pi}{2}} \tan(A+\theta)=\displaystyle \lim_{A\to \dfrac{\pi}{2}} \dfrac{\sin(A+\theta)}{\cos (A+\theta)}=\dfrac{\sin(\dfrac{\pi}{2}+\theta)}{\cos(\dfrac{\pi}{2}+\theta)}=\dfrac{\cos \theta}{-\sin \theta}=-\cot \theta$

DeepSea
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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}$$$=\lim_{t\to 0}\frac{1+(t)\tan\theta}{t-\tan \theta}$$ $$=\frac{1+0}{0-\tan \theta}$$ $$=-\frac{1}{\tan \theta}=-\cot \theta$$

Harish Chandra Rajpoot
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Using:

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

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

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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.

Steven Stadnicki
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