$\arctan x$ is defined as:
$$\arctan x = \frac{1}{\tan(x)} = \frac{1}{\frac{\sin(x)}{\cos(x)}}$$
if I now have $x = 0$ I should get:
$$\frac{1}{\frac{\sin(0)}{\cos(0)}} = \frac{1}{\frac{0}{1}} = \frac{1}{0} = \infty$$
but actually it is defined as $\mathrm{arctan}(0) = 0$.
What did I not get?
Bodo
$\endgroup$ 71 Answer
$\begingroup$You've hit on the big notational ambiguity in the notation for trigonometric functions. Usually $\tan^n(x)$ means $(\tan(x))^n$. This includes negative powers: $$ \tan^{-2}(x) = (\tan(x))^{-2} = \frac{1}{\tan(x)^2} $$ etc. The exception is the exponent $-1$. We reserve $\tan^{-1}(x)$ to mean the inverse to the tangent function. So $\tan^{-1}(0)$ is the angle whose tangent is zero, namely zero.
I try to avoid that ambiguity by never writing $\tan^{-1}$. I'll use $\arctan$ for the inverse tangent and $\cot$ for $\frac{1}{\tan}$. But the trap is so attractive it sometimes surpasses this measure. I see students write things like: $$ \color{red}{\arctan(x) = \tan^{-1}(x) = \frac{1}{\tan x} = \cot x} \qquad\text{(Wrong!)} $$ all the time. The first and third equal signs are correct, but not the one in the middle.
I myself did this on a high school calculus exam 25 years ago. Welcome to the club!
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