Question

If ${\cot }^{-1}(\frac{1-x^2}{2x})+{\cos }^{-1}(\frac{1-x^2}{1+x^2})=\frac{2\pi }{3},x>0$ , $x\ne 1$ then $x=$

• A. $\frac{1}{\sqrt{2}}$
• B. $\pm \frac{1}{\sqrt{2}}$
• C. $\pm \frac{1}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{3}}$

$co{t}^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{2\pi }{3}-co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
$co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\frac{2\pi }{3}-co{t}^{-1}\left(\frac{1-x^2}{2x}\right)$
take cos on both sides
Simplify, $\frac{1-x^2}{1+x^2}=-\frac{1}{2}\left(\frac{1-x^2}{1+x^2}\right)+\frac{\sqrt{3}}{2}\frac{2x}{1+x^2}\left[cos\right(A-B)=cosAcosB+sinAsinB]$
$\frac{2-2x^2-2\sqrt{3}x}{2(1+x^2)}=-\frac{1}{2}\frac{(1-x^2)}{(1+x^2)}$
$2-2x^2-2\sqrt{3}x=x^2-1$
$3x^2+2\sqrt{3}x-3=0$
$\therefore x=\frac{-2\sqrt{3}\pm \sqrt{12+36}}{6}$
$x=\frac{-2\sqrt{3}\pm 4\sqrt{3}}{6}$
$x=-\sqrt{3},\frac{\sqrt{3}}{3}$
$x=-\sqrt{3},\frac{1}{\sqrt{3}}$