Let the expression be $f(x) = (\sin^{-1} x)^2 + (\cos^{-1} x)^2$. We are given the domain $x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$. The maximum value is given in the form $\frac{m}{n} \pi^2$, and we need to find $m + n$.

Function Simplification

Using the identity $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$, we can write $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$. Let $y = \sin^{-1} x$. Substituting this into the expression for $f(x)$: $f(x) = y^2 + \left(\frac{\pi}{2} - y\right)^2$ $f(x) = y^2 + \left(\frac{\pi^2}{4} - \pi y + y^2\right)$ $f(x) = 2y^2 - \pi y + \frac{\pi^2}{4}$ Let $g(y) = 2y^2 - \pi y + \frac{\pi^2}{4}$.

Domain Range Mapping

The domain for $x$ is $\left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$. We need to find the corresponding range for $y = \sin^{-1} x$. The function $\sin^{-1} x$ is an increasing function. When $x = -\frac{\sqrt{3}}{2}$, $y = \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}$. When $x = \frac{1}{\sqrt{2}}$, $y = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$. So, the range for $y$ is $\left[-\frac{\pi}{3}, \frac{\pi}{4}\right]$.

Quadratic Maximum Value

We need to find the maximum value of the quadratic function $g(y) = 2y^2 - \pi y + \frac{\pi^2}{4}$ on the interval $y \in \left[-\frac{\pi}{3}, \frac{\pi}{4}\right]$. The graph of $g(y)$ is a parabola opening upwards (since the coefficient of $y^2$ is $2 > 0$). The vertex of the parabola occurs at $y = -\frac{b}{2a} = -\frac{(-\pi)}{2(2)} = \frac{\pi}{4}$. Since the vertex is at the right endpoint of the interval $\left[-\frac{\pi}{3}, \frac{\pi}{4}\right]$ and the parabola opens upwards, the function $g(y)$ is decreasing over this interval. Therefore, the maximum value of $g(y)$ occurs at the left endpoint, $y = -\frac{\pi}{3}$.

Result Calculation

Substitute $y = -\frac{\pi}{3}$ into $g(y)$: $g\left(-\frac{\pi}{3}\right) = 2\left(-\frac{\pi}{3}\right)^2 - \pi\left(-\frac{\pi}{3}\right) + \frac{\pi^2}{4}$ $g\left(-\frac{\pi}{3}\right) = 2\left(\frac{\pi^2}{9}\right) + \frac{\pi^2}{3} + \frac{\pi^2}{4}$ To add these fractions, find a common denominator, which is 36: $g\left(-\frac{\pi}{3}\right) = \frac{2 \times 4 \pi^2}{9 \times 4} + \frac{\pi^2 \times 12}{3 \times 12} + \frac{\pi^2 \times 9}{4 \times 9}$ $g\left(-\frac{\pi}{3}\right) = \frac{8\pi^2}{36} + \frac{12\pi^2}{36} + \frac{9\pi^2}{36}$ $g\left(-\frac{\pi}{3}\right) = \frac{(8 + 12 + 9)\pi^2}{36} = \frac{29\pi^2}{36}$. This maximum value is given as $\frac{m}{n} \pi^2$. Equating the two, we get $\frac{m}{n} = \frac{29}{36}$. We have $m = 29$ and $n = 36$. Check the condition $\gcd(m, n) = 1$: $\gcd(29, 36) = 1$, as 29 is prime. Finally, calculate $m + n$: $m + n = 29 + 36 = 65$.