Question

$co{s}^{-1}x=ta{n}^{-1}\frac{2x}{1+x^2}$, then:

• A. $xϵR$
  
• B. $-1x\le 1,x\ne 0$
  
• C. $0<x\le 1$
  
• D. $-1\le x<0$

Answer 1

The correct option is C $0<x\le 1$


Puttting $\theta =co{s}^{-1}x$ in RHS, we have
R.H.S $=ta{n}^{-1}\frac{\sqrt{1-x^2}}{x}$
$=ta{n}^{-1}\frac{\sqrt{1-co{s}^2\theta }}{cos\theta }(0\le \theta \le \pi )$
$=ta{n}^{-1}\frac{sin\theta }{cos\theta }=ta{n}^{-1}tan\theta$
$=\theta =co{s}^{-1}x$ when $-\frac{\pi }{2}<\theta <\frac{\pi }{2}$
i,e., $when0\le \theta \le \frac{\pi }{2}$
i.e., when $0<cos\theta \le 1$ i.e., $0<x\le 1$