Question

The greatest and least values of $({\sin }^{-1}x{)}^2+({\cos }^{-1}x{)}^2$ are respectively

• A. $\frac{{\pi }^2}{4}$ and 0
• B. $\frac{\pi }{2}$ and $\frac{-\pi }{2}$
• C. $\frac{5{\pi }^2}{4}$ and $\frac{{\pi }^2}{8}$
• D. $\frac{{\pi }^2}{4}$ and $\frac{{\pi }^2}{4}$

Answer 1

Answer: (C)

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 $2\left(\frac{{\pi }^2}{16}\right)\frac{{\pi }^2}{8}$
and the greatest value is
$2\left[{\left(\frac{-\pi }{2}-\frac{\pi }{4}\right)}^2+\frac{{\pi }^2}{16}\right]\frac{5{\pi }^2}{4}$