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
$2\left[\left(\frac{-\pi}{2}-\frac{\pi}{4}\right)^{2}+\frac{\pi^{2}}{16}\right] \frac{5 \pi^{2}}{4}$