$$\begin{gathered} \text{The maximum value of}\left({\sin ^-1x}^2\right)+\left({\cos ^-1x}^2\right)\text{in}x\in [\frac{-\sqrt{3}}{2},\frac{1}{\sqrt{2}}]\text{is }\frac{a{\pi }^2}{b}\text{, then}\mathrm{ a}+b\text{is} \\ \end{gathered}$$

The correct answer is 65.

Step-by-step Explanation:

We are asked to find the maximum value of the expression:

$f\left(x\right)=\left({\sin {}^{-1}x}^2\right)+\left({\cos {}^{-1}x}^2\right)$

for $x\in [-\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}]$.

We know the identity connecting inverse trigonometric functions:

$\sin {}^{-1}x+\cos {}^{-1}x=\frac{\pi }{2}$

Let $y=\sin {}^{-1}x$. Then $\cos {}^{-1}x=\frac{\pi }{2}-y$.

Since $x\in [-\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}}]$, the corresponding range of $y=\sin {}^{-1}x$ is:
$y\in \left[\sin {}^{-1}\right(-\frac{\sqrt{3}}{2}),\sin {}^{-1}(\frac{1}{\sqrt{2}}\left)\right]=[-\frac{\pi }{3},\frac{\pi }{4}]$

Now, rewrite $f\left(x\right)$ in terms of $y$:
$g\left(y\right)=y^2+{(\frac{\pi }{2}-y)}^2$

Expanding the expression:
$g\left(y\right)=y^2+\frac{{\pi }^2}{4}-\pi y+y^2=2y^2-\pi y+\frac{{\pi }^2}{4}$

This is a quadratic function in $y$ representing a parabola opening upwards. The vertex of this parabola is at:
$y=\frac{\pi }{4}$

Since the parabola opens upwards, the maximum value of $g\left(y\right)$ over the interval $y\in [-\frac{\pi }{3},\frac{\pi }{4}]$ occurs at the point furthest from the vertex $y=\frac{\pi }{4}$, which is $y=-\frac{\pi }{3}$.

Evaluating $g\left(y\right)$ at $y=-\frac{\pi }{3}$:
$g(-\frac{\pi }{3})=2{(-\frac{\pi }{3})}^2-\pi (-\frac{\pi }{3})+\frac{{\pi }^2}{4}$
$g(-\frac{\pi }{3})=2\left(\frac{{\pi }^2}{9}\right)+\frac{{\pi }^2}{3}+\frac{{\pi }^2}{4}$

Taking the common denominator, which is 36:
$g(-\frac{\pi }{3})=\frac{8{\pi }^2+12{\pi }^2+9{\pi }^2}{36}=\frac{29{\pi }^2}{36}$

Thus, the maximum value is given in the form:
$\frac{a{\pi }^2}{b}=\frac{29{\pi }^2}{36}$

Comparing the two expressions gives $a=29$ and $b=36$.

Therefore, the required value of $a+b$ is:
$a+b=29+36=65$