Question

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

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

Answer 1

The correct option is D $a$

Given that,

$2\sin h^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=log\left(\frac{1+x}{1-x}\right)$

$x=?$

We know that,

$\sin h^{-1}x=log(x+\sqrt{x^2+1})$

$2\sin h^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=log\left(\frac{1+x}{1-x}\right)$

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

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

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

${\left(\frac{a+1}{\sqrt{1-a^2}}\right)}^2=\frac{1+x}{1-x}$

$\frac{a+1}{1-a}=\frac{1+x}{1-x}$

$\frac{1+a}{1-a}=\frac{1+x}{1-x}$

Using componendo and dividendo rule,

$\frac{(1+a)+(1-a)}{(1+a)-(1-a)}=\frac{(1+x)+(1-x)}{(1+x)-(1-x)}$

$\frac{1+a+1-a}{1+a-1+a}=\frac{1+x+1-x}{1+x-1+x}$

$\frac{2}{2a}=\frac{2}{2x}$

$x=a$

Hence, this is the answer.