Question

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

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