Question

lf $2{\sinh }^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=\log \left(\frac{1+x}{1-x}\right)$ , then $x=$

• A. $2a$
• B. $3a$
• C. $4a$
• D. $a$

Answer 1

The correct option is D $a$

Formula $\to sin{h}^{-1}x=ln(x+\sqrt{x^2+1})$
$\therefore sin{h}^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=ln(\frac{a}{\sqrt{1-a^2}}+\sqrt{\frac{a^2}{1-a^2}+1})$
$=ln(\frac{a}{\sqrt{1-a^2}}+\frac{1}{\sqrt{1-a^2}})$
$=ln\left(\frac{\sqrt{a+1}}{\sqrt{1-a}}\right)$
$=ln(\frac{1+a}{1-a}{)}^{1/2}$
$sin{h}^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=\frac{1}{2}ln\left(\frac{1+a}{1-a}\right)$
$2sin{h}^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=ln\left(\frac{1+x}{1-x}\right)$ [Given ]
$\Rightarrow x=a$