Question

Which of the following is true in the interval of $x\in (0,1)$

• A. ${\sin }^{-1}x<{\tan }^{-1}x$
• B. ${\sin }^{-1}x\le {\tan }^{-1}x$
• C. ${\sin }^{-1}x>{\tan }^{-1}x$
• D. ${\sin }^{-1}x\ge {\tan }^{-1}x$

Answer 1

The correct option is C ${\sin }^{-1}x>{\tan }^{-1}x$

Let $f\left(x\right)={\sin }^{-1}x-{\tan }^{-1}x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{\sqrt{1-x^2}}-\frac{1}{1+x^2}$
$\because \sqrt{1-x^2}<1+x^2,\forall 0<x<1$
$\Rightarrow \frac{1}{\sqrt{1-x^2}}>\frac{1}{1+x^2}$
So, $f^{\prime }\left(x\right)>0$ for $x\in (0,1)$
$\Rightarrow f\left(x\right)$ strictly increases for $x\in (0,1)$
$\Rightarrow f\left(x\right)>f\left(0\right)$
$\Rightarrow {\sin }^{-1}x-{\tan }^{-1}x>0$