Find the greatest and least values of $(\sin^{-1}x)^2+(\cos^{-1}x)^2$

Let $\sin^{-1}x=a,\cos^{-1}x=b$

$a+b=\dfrac\pi2$

$$a^2+b^2=\left(\dfrac\pi2-b\right)^2+b^2=\dfrac{\pi^2}4+2\left(b-\dfrac\pi4\right)^2-2\left(\dfrac\pi4\right)^2$$

Again , $0\le b\le\pi\implies0\le\left(b-\dfrac\pi4\right)^2\le\left(\pi-\dfrac\pi4\right)^2$

By C-S $$\arcsin^2{x}+\arccos^2{x}=\frac{1}{2}(1^2+1^2)(\arcsin^2{x}+\arccos^2{x})\geq\frac{1}{2}(\arcsin{x}+\arccos{x})^2=\frac{\pi^2}{8}.$$ The equality occurs for $x=\frac{1}{\sqrt2}$, which says that we got a minimal value.

In another hand, for $x\geq0$ we obtain: $$\arcsin^2{x}+\arccos^2{x}\leq\left(\arcsin{x}+\arccos{x}\right)^2=\frac{\pi^2}{4}.$$ The equality occurs for $x=0$.

For $x<0$ let $t=-\arcsin{x}$.

Thus, $0<t\leq\frac{\pi}{2}$, $\arccos{x}=\pi-\arccos(-x)=\pi-\left(\frac{\pi}{2}-\arcsin(-x)\right)=\frac{\pi}{2}+t$ and $$\arcsin^2{x}+\arccos^2{x}=t^2+\left(\frac{\pi}{2}+t\right)^2\leq\frac{5\pi^2}{4}$$ with equality for $x=-1$, which says that $\frac{5\pi^2}{4}$ is a maximal value.

You can do it with calculus: if $f(x)=(\arcsin x)^2+(\arccos x)^2$, then $$ f'(x)=\frac{2\arcsin x}{\sqrt{1-x^2}}-\frac{2\arccos x}{\sqrt{1-x^2}} $$ (for $-1<x<1$). The derivative vanishes for $\arcsin x=\arccos x$, that is, $x=1/\sqrt{2}$. Since $$ f(1/\sqrt{2})=\frac{\pi^2}{8} \qquad f(-1)=\frac{5\pi^2}{4} \qquad f(1)=\frac{\pi^2}{4} $$ you can easily draw the conclusion.