Question

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

• A. $X\in R$
• B. $-\le x\le 1,x\ne 0$
• C. $0<x\le 1$
• D. $Noneofthese$

Answer 1

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

Putting $\theta =co{s}^{-1}x$ in R.H.S., 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., when $0\le \theta <\frac{\pi }{2}$
i.e., when $0<cos\theta \le 1$ i.e., $0<x\le 1$.