Question

The solution of the equation ${\sin }^{-1}\sqrt{1-x^2}+{\cos }^{-1}x={\cot }^{-1}\frac{\sqrt{1-x^2}}{x}-{\sin }^{-1}x$ is

• A. $[-1,1]-\left\{0\right\}$
• B. $(0,1)\cup \{-1\}$
• C. $[-1,0)\cup \left\{1\right\}$
• D. $[-1,1]$

Answer 1

The correct option is C $[-1,0)\cup \left\{1\right\}$

$si{n}^{-1}\sqrt{1-x^2}+co{s}^{-1}x=co{t}^{-1}\frac{\sqrt{1-x^2}}{x}-si{n}^{-1}x$
For $si{n}^{-1}\sqrt{1-x^2}$ $-1\le \sqrt{1-x^2}\le 1\Rightarrow x\le 0$ ...(1)
And for $co{s}^{-1}x-1\le x\le 1$ ...(2)
From (1) and (2)
$xϵ\{-1,0\}$ ...(3)
Now,
$si{n}^{-1}\sqrt{1-x^2}+co{s}^{-1}x=co{t}^{-1}\frac{\sqrt{1-x^2}}{x}-si{n}^{-1}x\Rightarrow si{n}^{-1}\sqrt{1-x^2}+si{n}^{-1}\sqrt{1-x^2}=si{n}^{-1}x-si{n}^{-1}x$
$\Rightarrow 2si{n}^{-1}\sqrt{1-x^2}=0\Rightarrow \sqrt{1-x^2}=0\Rightarrow x=1$ ...(4)
From (3) and (4)
$xϵ\{(-1,0)\cup \left\{1\right\}\}$