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


  $$
(\sin^{-1}x)^2+(\cos^{-1}x)^2
$$


My Attempt:

$$
\frac{-\pi}{2}\leq\sin^{-1}x\leq \frac{\pi}{2}\quad\&amp;\quad0\leq\cos^{-1}x\leq\pi\\(\sin^{-1}x)^2\leq\frac{\pi^2}{4}\quad\&amp;\quad(\cos^{-1}x)^2\leq\pi^2\\
0\leq(\sin^{-1}x)^2+(\cos^{-1}x)^2\leq\frac{\pi^2}{4}+\pi^2=\frac{5\pi^2}{4}
$$

Here, I can see the upper limit is $\frac{5\pi^2}{4}$ which is fine. But, $0$ is one lower limit not the lower limit.

Why am I not getting the lower limit in my approach ?

How do I approach similar problems involving max and min, when you don't get the lower or upper limits ?
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 , $0\le b\le\pi\implies0\le\left(b-\dfrac\pi4\right)^2\le\left(\pi-\dfrac\pi4\right)^2$
Given, differentiate $f(x)$ with respect to $x$ now, $f'(x) = 0$ then, We also know, domain of  and  $∈ [ -1 1] ∴$ find $f(x)$ at $x = 1$ , $-1$ and $\frac{1}{\sqrt{2}}$ 

$f(1) = (sin^{-1}(1))^2 + (cos^{-1}(1))^2 = \frac{π^2}{4} + 0 = \frac{π^2}{4}$ $f(-1) = (sin^{-1}(-1))^2 + (cos^{-1}(-1))^2 = \frac{π^2}{4} + π^2 = \frac{5π^2}{4}$ $f(\frac{1}{\sqrt{2}}) = (sin^{-1}(\frac{1}{\sqrt{2}}))^{2} + (cos^{-1}(\frac{1}{\sqrt{2}}))^2 = \frac{π^2}{16} + \frac{π^2}{16} = \frac{π^2}{8}$

Hence, maximum value of $f(x) = \frac{5π^2}{4}$ Minimum value of $f(x) = \frac{π^2}{8}$