Question

If $\left|{\cos }^{-1}\frac{1-x^2}{1+x^2}\right|<\frac{\pi }{3}$, then

• A. $x\in [-\frac{1}{3},\frac{1}{\sqrt{3}}]$
• B. $x\in (-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$
• C. $x\in (0,\frac{1}{\sqrt{3}})$
• D. none of these

Answer 1

Answer: (B)

$x\in (-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$

$\left|{\cos }^{-1}\right(\frac{1-x^2}{1+x^2}\left)\right|<\frac{\pi }{3}$

Simplifying the above inequality by taking cosine on both the sides, we get
$\frac{1-x^2}{1+x^2}>\frac{1}{2}$ ...(cos(x) is decreasing function)
$\Rightarrow 2-2x^2>1+x^2$
$\Rightarrow 1>3x^2$
$\Rightarrow x^2<\frac{1}{3}$

Hence $x\in (\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}})$