Mathematics

# $\displaystyle\dfrac{e^{2x}-1}{e^{2x}+1}$ in terms of hyperbolic function is

##### SOLUTION
$\dfrac{{e}^{2x}-1}{{e}^{2x}+1}$
Remove ${e}^{x}$ from both the numerator and denominator.
$=\dfrac{{e}^{x}\left({e}^{x}-{e}^{-x}\right)}{{e}^{x}\left({e}^{x}+{e}^{-x}\right)}$
$=\dfrac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}$
$\tan{hx}$ which is a hyperbolic function

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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