Question

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

• A. $\frac{{\pi }^2}{4}and0$
• B. $\frac{\pi }{2}and-\frac{\pi }{2}$
• C. $\frac{5{\pi }^2}{4}and\frac{{\pi }^2}{8}$
• D. $\frac{{\pi }^2}{4}and\frac{-\pi }{4}$

Answer 1

Answer: (C)

$\frac{5{\pi }^2}{4}and\frac{{\pi }^2}{8}$

$\begin{array}{cc}\text{let}& {\sin }^{-1}x=0\\ & =x=\sin \theta \end{array}$

${\cos }^{-1}x$ = $\pi /2-\theta$

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

$o^2+{(\frac{\pi }{2}-0)}^2$

thus max value is = $\frac{5{\pi }^2}{4}$

min value is =$\frac{{\pi }^2}{8}$