Question

Which of the following, satisfy the equation $2{\cos }^{-1}x={\cot }^{-1}\left(\frac{2x^2-1}{\sqrt{4x^2-4x^4}}\right)$

• A. $(-1,0)$
• B. $(0,1)$
• C. $\left(-\frac{1}{2},\frac{1}{\sqrt{2}}\right)$
• D. $[-1,1]$

Answer 1

Answer: (B)

$2\cdot {\cos }^{-1}x={\cot }^{-1}\left(\frac{2x^2-1}{2|x|\sqrt{1-x^2}}\right)$
Let ${\cos }^{-1}x=\theta$
$2\theta ={\cot }^{-1}\left(\frac{\cos 2\theta }{2|\cos \theta |\sin \theta }\right)$
Case I : If $\cos \theta >0,x>0\Rightarrow 0<x<1$ then
$2\theta ={\cot }^{-1}\cot 2\theta =2\theta$ (identity)
Case II : $\cos \theta <0$, which not satisfy the equation