AP EAMCET · Maths · Inverse Trigonometric Functions
Let \(\mathrm{x} \in \mathbf{R}\) and \(|\mathrm{x}| < 1\). Then \(\tanh ^{-1} x=\)
- 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 \(\frac{1}{2} \log \left(x-\sqrt{1-x^2}\right)\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2} \log \left(\frac{1+x}{1-x}\right)\)
Step-by-step Solution
Detailed explanation
Let \(y=\tan h^{-1} x\) \(\begin{aligned} & x=\tan h y \\ & x=\frac{e^y-e^{-y}}{e^y+e^{-y}}\end{aligned}\) \(\begin{aligned} & x e^y+e^{-y} x=e^y-e^{-y} \\ & (1-x) e^y=(1+x) e^{-y} \\ & e^{2 y}=(1-x)=(1+x)\end{aligned}\)…
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