Question

If $-1<x<0$, then ${\cos }^{-1}x$ is equal to

• A. ${\sec }^{-1}\frac{1}{x}$
• B. $\pi -{\sin }^{-1}\sqrt{1-x^2}$
• C. $\pi +{\tan }^{-1}\frac{\sqrt{1-x^2}}{x}$
• D. ${\cot }^{-1}\frac{x}{\sqrt{1-x^2}}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A ${\sec }^{-1}\frac{1}{x}$
B $\pi +{\tan }^{-1}\frac{\sqrt{1-x^2}}{x}$
C $\pi -{\sin }^{-1}\sqrt{1-x^2}$
D ${\cot }^{-1}\frac{x}{\sqrt{1-x^2}}$

Let $x=cos\theta$ , $\frac{\pi }{2}<\theta <\pi$
$co{s}^{-1}x=\theta$ , $-1<x<0$
The range of $se{c}^{-1}$ function is $[0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi ]$.
$se{c}^{-1}\left(\frac{1}{x}\right)=se{c}^{-1}\left(\frac{1}{cos\theta }\right)$
$\Rightarrow se{c}^{-1}\left(\frac{1}{x}\right)=se{c}^{-1}\left(sec\theta \right)=\theta$
Hence, option A is correct.
$si{n}^{-1}\sqrt{1-x^2}=si{n}^{-1}\sqrt{1-co{s}^2\theta }$
$\Rightarrow si{n}^{-1}\sqrt{1-x^2}=si{n}^{-1}\left(|sin\theta |\right)$
$|sin\theta |>0$ for $-1<x<0$
$\Rightarrow si{n}^{-1}\sqrt{1-x^2}=si{n}^{-1}(-sin\theta )$
$\Rightarrow si{n}^{-1}\sqrt{1-x^2}=si{n}^{-1}\left(sin(\pi -\theta )\right)$
$\Rightarrow si{n}^{-1}\sqrt{1-x^2}=\pi -\theta$ (The range of sin inverse function is $[\frac{-\pi }{2},\frac{\pi }{2}]$
Hence, option B is correct.
$ta{n}^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)=ta{n}^{-1}\left(\frac{\sqrt{1-co{s}^2\theta }}{cos\theta }\right)$
$\Rightarrow ta{n}^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)=ta{n}^{-1}\left(\frac{|sin\theta |}{cos\theta }\right)$
$\Rightarrow ta{n}^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)=ta{n}^{-1}\left(tan\theta \right)$
$\Rightarrow ta{n}^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)=ta{n}^{-1}\left(tan\right(\theta -\pi ))$
$\Rightarrow ta{n}^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)=\theta -\pi$ (The range of tan inverse function is $(\frac{-\pi }{2},\frac{\pi }{2}))$
Hence, option C is correct.
Similarly,
$co{t}^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)=co{t}^{-1}\left(cot\theta \right)=\theta$ (The range of cot inverse function is $(0,\pi )$ )
Hence, option D is also correct