Question

If $a\le {\sin }^{-1}x+{\cos }^{-1}x+{\tan }^{-1}x\le b$, then

• A. $a=0,b=\pi$
• B. $a=0,b=\frac{\pi }{2}$
• C. $a=\frac{\pi }{2},b=\pi$
• D. None of these

Answer 1

The correct option is A $a=0,b=\pi$

$\because \frac{\pi }{2}={\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}.....\left(i\right)$
and $-\frac{\pi }{2}<{\tan }^{-1}x<\frac{\pi }{2}.....\left(ii\right)$
Adding eqs. (i) and (ii), we get
$\frac{\pi }{2}-\frac{\pi }{2}<{\sin }^{-1}x+{\cos }^{-1}x+{\tan }^{-1}x<\frac{\pi }{2}+\frac{\pi }{2}$
$\Rightarrow 0<{\sin }^{-1}x+{\cos }^{-1}x+{\tan }^{-1}x<\pi$
Comparing it with given condition, we get
$a=0,b=\pi$.