Question

If $co{t}^{-1}(\frac{1}{x+1})+co{t}^{-1}(\frac{1}{x-1})=$$ta{n}^{-1}3x-ta{n}^{-1}x$ ,
then $x=$

• A. $\pm 1/2$
• B. $-1,\pm 1/3$
• C. $2,\pm 1$
• D. $-1,\pm 1/2$

Answer 1

The correct option is A $\pm 1/2$

$co{t}^{-1}\left(\frac{1}{x+1}\right)+co{t}^{-1}\left(\frac{1}{x-1}\right)=ta{n}^{-1}(x+1)+ta{n}^{-1}(x-1)$
$=ta{n}^{-1}\left(\frac{x+1+x-1}{1-(x^2-1)}\right)$
$=ta{n}^{-1}\left(\frac{2x}{2-x^2}\right)$ 1
and
$ta{n}^{-1}3x-ta{n}^{-1}x=ta{n}^{-1}\left(\frac{3x-x}{1+3x^2}\right)=ta{n}^{-1}\left(\frac{2x}{1+3x^2}\right)$ 2
equation 1 = 2
$ta{n}^{-1}\frac{2x}{1+3x^2}=ta{n}^{-1}\left(\frac{2x}{2-x^2}\right)$
$1+3x^2=2-x^2$
$4x^2=1$
$x^2=\frac{1}{4}$
$\Rightarrow x=\pm \frac{1}{2}$