Let the maximum value of $$\left(\sin^{-1}x\right)^2+\left(\cos^{-1}x\right)^2$$ for $$x\epsilon \left[-\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}\right]$$ be $$\frac{m}{n}\pi^{2}$$, where gcd
(m, n) = l. Then m + n is equal to ____________

We begin by noting that we need to find the maximum value of $$(\sin^{-1}x)^2 + (\cos^{-1}x)^2$$ for $$x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$$.

Introducing the substitution $$\theta = \sin^{-1}x$$ leads to $$\cos^{-1}x = \frac{\pi}{2} - \theta$$.

This yields the function $$f(\theta) = \theta^2 + \left(\frac{\pi}{2} - \theta\right)^2 = \theta^2 + \frac{\pi^2}{4} - \pi\theta + \theta^2 = 2\theta^2 - \pi\theta + \frac{\pi^2}{4}$$.

When $$x = -\frac{\sqrt{3}}{2}$$ we have $$\theta = \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3}$$.

When $$x = \frac{1}{\sqrt{2}}$$ we obtain $$\theta = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$$.

Hence $$\theta \in \left[-\frac{\pi}{3}, \frac{\pi}{4}\right]$$.

Taking the derivative gives $$f'(\theta) = 4\theta - \pi = 0 \Rightarrow \theta = \frac{\pi}{4}$$.

Since the parabola opens upward, the minimum at $$\theta = \frac{\pi}{4}$$ is at the right endpoint, so the maximum occurs at the endpoint farthest from the vertex.

Evaluating at $$\theta = -\frac{\pi}{3}$$ gives $$f\left(-\frac{\pi}{3}\right) = 2 \cdot \frac{\pi^2}{9} + \frac{\pi^2}{3} + \frac{\pi^2}{4} = \frac{2\pi^2}{9} + \frac{\pi^2}{3} + \frac{\pi^2}{4} = \pi^2\left(\frac{2}{9} + \frac{1}{3} + \frac{1}{4}\right) = \pi^2\left(\frac{8 + 12 + 9}{36}\right) = \frac{29}{36}\pi^2$$.

Evaluating at $$\theta = \frac{\pi}{4}$$ yields $$f\left(\frac{\pi}{4}\right) = 2 \cdot \frac{\pi^2}{16} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}$$.

Therefore, the maximum value is $$\frac{29}{36}\pi^2$$, occurring at $$x = -\frac{\sqrt{3}}{2}$$.

This value can be written as $$\frac{m}{n}\pi^2 = \frac{29}{36}\pi^2$$ with $$\gcd(29,36) = 1$$.

It follows that $$m + n = 29 + 36 = 65$$.

The answer is 65.