Is $ \tan(\arctan(\frac{1}{x})) $ equal to infinity?

Question

I am solving the equation below: $$ S = \pi (4 - (p . \tan(\arctan(\frac{1}{p}))^2) $$ My solution for solving $ \tan(\arctan(\frac{1}{p})) $ is: $$ \tan(\arctan(\frac{1}{p})) = \tan(\frac{\pi}{2} - \arctan(p)) = \frac{\tan(\frac{\pi}{2}) - \tan(\arctan(p))}{ 1 + \tan(\frac{\pi}{2}) .\tan(\arctan(p))} = $$ $$ \frac{∞ - p}{1 + ∞ . p} = \frac{∞}{∞} $$ I know it is wrong! Please determine my mistake(s) and show me how to solve such equations. thanks!

Answer 1

I'm not a trig identity wizard but I can tell you that $\tan(\arctan(x))=x$ as the definition of arctan is the inverse of tan. So you don't need to do all those manipulations to solve for $\tan(\arctan(\frac{1}{p}))$ as that is just $\frac{1}{p}$. This is consistent with your answer as what you got is undefined and could be any value.