Question

If $2\tan h^{-1}x=\log y$, then the value of $y$ in the terms of $x$ is

• A. $2x$
• B. $\frac{2x}{1-x^2}$
• C. $x^2$
• D. $\left(\frac{1+x}{1-x}\right)$

Answer 1

Answer: (D)

$\left(\frac{1+x}{1-x}\right)$

Given that,

$2\tan h^{-}1x=\log y$

$\tan h^{-}1x=\frac{1}{2}\log y$

we know that,

$\tan h^{-}1x=\frac{1}{2}\log y$

then,

$\tan h^{-}1x=\frac{1}{2}\log y$

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

$\Rightarrow \log \left(\frac{1+x}{1-x}\right)=\log y$

$\Rightarrow \frac{1+x}{1-x}=y$

$\Rightarrow y=\frac{1+x}{1-x}$

Hence this is the answer