Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ The following identity is true for any given $x \in [-1,1]$:
$$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$
But I don't know how to explain it.
I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?
$$\int^{x}_{C1}\frac{1\cdot dx}{\sqrt{1-x^{2}}} + \int^{x}_{C2}\frac{-1 \cdot dx}{\sqrt{1-x^{2}}} =\\
 \arcsin(x) - \arcsin(C1) + \arccos(x) - \arccos(C2) = 0 \\
 \text{while } \arcsin(C1) + \arccos(C2) = \frac{\pi}{2}$$
I can't find the right words to explain why this is true?

Edit #1 (25 Jan, 20:10 UTC):
The following is a truth clause:
$$
\begin{array}{ll}
\frac{d}{dx}(\arcsin(x) + \arccos(x)) = \frac{d}{dx}\frac{\pi}{2} \\
\\
\frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} = 0
\end{array}
$$
By integrating the last equation, using the limits $k$ (a constant) and $x$ (variable), I get the following:
$$
\begin{array}{ll}
\int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\
\\
\arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = m  \text{ (m is a constant)}\\
\\
\arcsin(x) + \arccos(x) = m + \arcsin(k) + \arccos(k) \\
\\
\text{Assuming that } A = m + \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1]
\end{array}
$$
Using Calculus, why is that true for every $x \in [-1,1]$?

Edit #2:
A big mistake of mine was to think that $\int^x_k0 = m \text{ (m is const.)}$, but that isn't true for definite integrals.
Thus the equations from "Edit #1" should be as follows:
$$
\begin{array}{ll}
\int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\
\\
\arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = 0\\
\\
\arcsin(x) + \arccos(x) = \arcsin(k) + \arccos(k) \\
\\
A = \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1]
\end{array}
$$

By definition, $\arcsin(x)$ is the angle $\alpha$  such that $\sin(\alpha) = x$ and $-\pi/2 \le \alpha \le \pi/2$, while $\arccos(x)$ is the angle $\beta$ such that $\cos(\beta) = x$ and $0 \le \beta \le \pi$.  Since  $-\pi/2 \le \alpha \le \pi/2$, $\cos(\alpha) \ge 0$, so we have $\cos(\alpha) = \sqrt{1 - x^2}$.  Similarly $\sin(\beta) = \sqrt{1-x^2}$.  Now
$$\eqalign{-\pi/2 &\le \arcsin(x) + \arccos(x) = \alpha + \beta  \le 3 \pi/2 \cr\sin(\alpha + \beta) &= \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta) = 
x^2 + 1 - x^2 = 1\cr
\cos(\alpha + \beta) &= \cos(\alpha) \cos(\beta) - \sin(\alpha)\sin(\beta) =  \sqrt{1-x^2} x - x \sqrt{1-x^2} = 0}$$
and the only angle in this interval with that sine and cosine is $\pi/2$.
