Prove that $f(x)=\arcsin(x)+\arccos(x)$ is constant function and it is $f(x)= \frac{\pi}{2}$

Notice that $\arcsin y$ is the solution of $y=\sin x$ over $x\in [-\frac{\pi}2,\frac{\pi}2]$ and $\arccos y$ is the solution of $y=\cos x$ over $x\in [0,\pi]$. So if we have $x_0=\arcsin y$, then we have $$ y=\sin x_0 \implies y=\cos\left(\frac \pi 2-x_0\right) $$ and $\frac \pi 2-x_0\in [0,\pi]$. This gives $\arccos y = \frac \pi 2 -x_0=\frac\pi 2 - \arcsin y$ as wanted.

We have \begin{align} \cos f(x) &= \cos(\overbrace{\arcsin(x)}^{\in\left[-\frac\pi2, \frac\pi2\right]})\cos(\arccos(x)) - \sin(\arcsin(x))\sin(\overbrace{\arccos(x)}^{\in[0,\pi]})\\ &= \sqrt{1-\sin^2(\arcsin(x))}\cos(\arccos(x)) - \sin(\arcsin(x))\sqrt{1-\cos^2(\arccos(x))}\\ &= \sqrt{1-x^2}\cdot x - x \sqrt{1-x^2}\\ &= 0 \end{align} so the image of $f : [-1,1] \to \mathbb{R}$ is contained within $\left\{\frac\pi2 + k\pi : k \in \mathbb{Z}\right\}$. Since $f$ is continuous and $[-1,1]$ is connected, we conclude that image of $f$ must be a singleton so $f$ is constant.

Plugging in $x = 0$ yields $f \equiv \frac\pi2$.

Notice that $f'(x) = 0$ for all $x \in \mathbb{R}$. So $f(x) = c$ for a constant $c \in \mathbb{R}.$ Since $f(0) = \frac{\pi}{2}$, one has $c = \frac{\pi}{2}.$ Hence the conclusion.