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Question: Answered & Verified by Expert
Let $\mathrm{x} \in \mathbf{R}$ and $|\mathrm{x}| < 1$. Then $\tanh ^{-1} x=$
MathematicsInverse Trigonometric FunctionsAP EAMCETAP EAMCET 2022 (05 Jul Shift 2)
Options:
  • A $\frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$
  • B $\frac{1}{2} \log \left(\frac{1-x}{1+x}\right)$
  • C $\frac{1}{2} \log \left(x+\sqrt{1-x^2}\right)$
  • D $\frac{1}{2} \log \left(x-\sqrt{1-x^2}\right)$
Solution:
1599 Upvotes Verified Answer
The correct answer is: $\frac{1}{2} \log \left(\frac{1+x}{1-x}\right)$
Let $y=\tan h^{-1} x$
$\begin{aligned} & x=\tan h y \\ & x=\frac{e^y-e^{-y}}{e^y+e^{-y}}\end{aligned}$
$\begin{aligned} & x e^y+e^{-y} x=e^y-e^{-y} \\ & (1-x) e^y=(1+x) e^{-y} \\ & e^{2 y}=(1-x)=(1+x)\end{aligned}$
$\begin{aligned} & e^{2 y}=\left(\frac{1+x}{1-x}\right) \\ & \log e^{2 y}=\log \left(\frac{1+x}{1-x}\right)\end{aligned}$
$\begin{aligned} & 2 y=\log \left(\frac{1+x}{1-x}\right) \\ & y=\frac{1}{2} \log \left(\frac{1+x}{1-x}\right)\end{aligned}$

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