Question

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

• A. $a$
• B. $\frac{1}{a}$
• C. $\sqrt{1-a^2}$
• D. $\frac{1}{\sqrt{1-a^2}}$

Answer 1

The correct option is A $a$

As we know that

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

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

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

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

$=\log (\frac{(1+a{)}^2}{(1-a)(1+a)})$

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

But given $2{\sinh }^{-1}(\frac{a}{\sqrt{1-a^2}})=\log (\frac{1+x}{1-x})$

So $x=a$