Let the maximum value of (sin^-1x)^2 + (cos^-1x)^2$(\sin^{-1}x)^{2} + (\cos^{-1}x)^{2}$ for x in left[-fracsqrt32, frac1sqrt2right]$x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$ be \frac{m}{n}\pi^{2}, where gcd(m, n) = 1$\gcd(m, n) = 1$. Then m + n$m + n$ is equal to ____.

### Related Formula sin^-1x + cos^-1x = fracpi2$$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$$ a^2 + b^2 = (a+b)^2 - 2ab$$a^2 + b^2 = (a+b)^2 - 2ab$$ ### Core Logic Let y = (sin^-1x)^2 + (cos^-1x)^2$y = (\sin^{-1}x)^2 + (\cos^{-1}x)^2$. y = (sin^-1x + cos^-1x)^2 - 2sin^-1x cos^-1x$$y = (\sin^{-1}x + \cos^{-1}x)^2 - 2\sin^{-1}x \cos^{-1}x$$ y = fracpi^24 - 2sin^-1x left(fracpi2 - sin^-1xright)$$y = \frac{\pi^2}{4} - 2\sin^{-1}x \left(\frac{\pi}{2} - \sin^{-1}x\right)$$ y = 2(sin^-1x)^2 - pi sin^-1x + fracpi^24$$y = 2(\sin^{-1}x)^2 - \pi \sin^{-1}x + \frac{\pi^2}{4}$$ Complete the square: y = 2left( sin^-1x - fracpi4 right)^2 - fracpi^28 + fracpi^24 = 2left( sin^-1x - fracpi4 right)^2 + fracpi^28$$y = 2\left( \sin^{-1}x - \frac{\pi}{4} \right)^2 - \frac{\pi^2}{8} + \frac{\pi^2}{4} = 2\left( \sin^{-1}x - \frac{\pi}{4} \right)^2 + \frac{\pi^2}{8}$$ ### Step 1: Bound the Function Given domain x in left[ -fracsqrt32, frac1sqrt2 right]$x \in \left[ -\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}} \right]$. The range of t = sin^-1x$t = \sin^{-1}x$ for this domain is left[ -fracpi3, fracpi4 right]$\left[ -\frac{\pi}{3}, \frac{\pi}{4} \right]$. So the expression is f(t) = 2left( t - fracpi4 right)^2 + fracpi^28$f(t) = 2\left( t - \frac{\pi}{4} \right)^2 + \frac{\pi^2}{8}$. ### Step 2: Find the Maximum The function f(t)$f(t)$ is a downward-opening distance squared logic? No, leading coefficient is positive, it's an upward parabola. Max value occurs at the boundary furthest from the vertex t = fracpi4$t = \frac{\pi}{4}$. The boundaries are -fracpi3$-\frac{\pi}{3}$ and fracpi4$\frac{\pi}{4}$. The distance from -fracpi3$-\frac{\pi}{3}$ to fracpi4$\frac{\pi}{4}$ is greatest. At t = -fracpi3$t = -\frac{\pi}{3}$: textMax = 2left( -fracpi3 - fracpi4 right)^2 + fracpi^28 = 2left( -frac7pi12 right)^2 + fracpi^28$$\text{Max} = 2\left( -\frac{\pi}{3} - \frac{\pi}{4} \right)^2 + \frac{\pi^2}{8} = 2\left( -\frac{7\pi}{12} \right)^2 + \frac{\pi^2}{8}$$ = 2left( frac49pi^2144 right) + fracpi^28 = frac49pi^272 + frac9pi^272 = frac58pi^272 = frac29pi^236$$= 2\left( \frac{49\pi^2}{144} \right) + \frac{\pi^2}{8} = \frac{49\pi^2}{72} + \frac{9\pi^2}{72} = \frac{58\pi^2}{72} = \frac{29\pi^2}{36}$$ ### Step 3: Calculate Required Value Here m = 29, n = 36$m = 29, n = 36$. Check gcd(29, 36) = 1$\gcd(29, 36) = 1$. m + n = 29 + 36 = 65$m + n = 29 + 36 = 65$. ### Pattern Recognition Expressions shaped like f(x)^2 + g(x)^2$f(x)^2 + g(x)^2$ where f(x)+g(x) = C$f(x)+g(x) = C$ will always map to a simple parabola. Analyze strictly based on vertex distance in the restricted f(x)$f(x)$ domain bounds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Inverse Trigonometric Functions Class 11 Maths: Quadratic Equations