inverse trigonometric equation $\tan^{-1}{x}+\cot^{-1}{x}=\frac{\pi}{2}$ I have problem with showing that $\displaystyle \tan^{-1}{x}+\cot^{-1}{x}=\frac{\pi}{2}$ I think there have to be used formula: $\displaystyle \tan(\alpha+\beta)=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}} $ but I don't know how to apply it and yet I don't know whether it's true that $\displaystyle\cot^{-1}{x}=\frac{1}{\tan^{-1}{x}}$ ? 
 A: Let $y = \arctan x$, then
\begin{align}
\tan y&=x\\
\frac{1}{\tan y}&=\frac{1}{x}\\
\cot y&=\frac{1}{x}\\
\tan\left(\frac\pi2-y\right)&=\frac{1}{x}\\
\frac\pi2-y&=\arctan\left(\frac{1}{x}\right)\\
\frac\pi2-\arctan x&=\text{arccot}\ x\\
\large\color{blue}{\arctan x+\text{arccot}\ x}&\color{blue}{=\frac\pi2}.\qquad\qquad\blacksquare
\end{align}
Here is the link for the proof of $\ \arctan\left(\frac1x\right) = \text{arccot}(x)$.
A: Note $\pi/2$ is $90^\circ$. Consider the rectangle in the following diagram:

A: The correct identity should be
$$\cot^{-1}{x}=\tan^{-1}\left(\frac{1}{x}\right)$$
The equation becomes
$$\tan^{-1}{x}+\tan^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}$$
We will take the tangent then the inverse tangent of the LHS to get
\begin{align}
\tan^{-1}\left(\tan\left(\tan^{-1}{x}+\tan^{-1}\left(\frac{1}{x}\right)\right)\right)
&=\tan^{-1}\left(\frac{x+1/x}{1-1}\right)
\end{align}
Clearly the term inside is not defined, and this occurs at $\dfrac{(2n+1)\pi}{2}$. But since the domain of the function $\tan^{-1}{x}$ is restriced to $-\dfrac{\pi}{2}\le 0 \le \dfrac{\pi}{2}$, we can say
$$\tan^{-1}{x}+\tan^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}$$
A: A calculus proof:
Let $f(x)=\arctan(x)+\textrm{arcot}(x)$ for all $x\in {\mathbb R}$. The function $f$ is differentiable and $f'(x)=\frac{1}{1+x^2}+\frac{-1}{1+x^2}=0$, so $f$ is constant. The value of the constant is $f(0)=\arctan(0)+\textrm{arccot}(0)=0+\frac{\pi}{2}$ (remember that $\tan(0)=0$ and $\cot(\frac{\pi}{2})=0$), so $\arctan(x)+\textrm{arccot}(x)=\frac{\pi}{2}$ for all $x\in {\mathbb R}$.
Note: I prefer the notation $\arctan$ over $\tan^{-1}$ since it helps to avoid mistakes like $\cot^{-1}(x)=\frac{1}{\tan^{-1}(x)}$.

Edit: the precalculus tag was added while I was writing my answer. 
