Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

We have, ${\left({\sin }^{-1}x\right)}^2+{\left({\cos }^{-1}x\right)}^2$ $={\left({\sin }^{-1}x+{\cos }^{-1}x\right)}^2-2{\sin }^{-1}x\cdot {\cos }^{-1}x$ $=\frac{{\pi }^2}{4}-2{\sin }^{-1}x\left(\frac{\pi }{2}-{\sin }^{-1}x\right)$ $=\frac{{\pi }^2}{4}-\pi {\sin }^{-1}x+2{\left({\sin }^{-1}x\right)}^2$ $=2\left[{\left({\sin }^{-1}x\right)}^2-\frac{\pi }{2}{\sin }^{-1}x+\frac{{\pi }^2}{8}\right]$ $=2\left[{\left({\sin }^{-1}x-\frac{\pi }{4}\right)}^2+\frac{{\pi }^2}{16}\right]$ Thus, the least value is $2\left(\frac{{\pi }^2}{16}\right)$ i.e. $\frac{{\pi }^2}{8}$ and the greatest value is $2\left[{\left(\frac{-\pi }{2}-\frac{\pi }{4}\right)}^2+\frac{{\pi }^2}{16}\right]$ i.e. $\frac{5{\pi }^2}{4}.$