Let the maximum value of $(\sin^{-1}x)^2 + (\cos^{-1}x)^2$ for $x \in \left[-\frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}\right]$ be $\frac{m\pi^2}{n}$, where $\gcd(m,n)=1$. Then m+n is equal to ______.
🎥 Video solution / Text Solution of this question is given below:
$(\sin^{-1}x)^2 + (\cos^{-1}x)^2$
$= (\sin^{-1}x + \cos^{-1}x)^2 - 2\sin^{-1}x\cos^{-1}x$
$= \frac{\pi^2}{4} - 2(\sin^{-1}x)\left(\frac{\pi}{2} - \sin^{-1}x\right)$
$= 2\left(\sin^{-1}x - \frac{\pi}{4}\right)^2 + \frac{\pi^2}{8}$
Maximum occurs at $\sin^{-1}x = -\frac{\pi}{3}$
$\Rightarrow 2\left(\frac{\pi}{3} + \frac{\pi}{4}\right)^2 + \frac{\pi^2}{8}$
$= 2\left(\frac{7\pi}{12}\right)^2 + \frac{\pi^2}{8}$
$= \frac{49\pi^2}{72} + \frac{9\pi^2}{72} = \frac{29\pi^2}{36}$
$\Rightarrow m = 29,; n = 36$
$\Rightarrow m+n = 65$