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
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