The value of (\frac{e^{2\theta} - 1}{e^{2\theta} + 1}) is. | MATH - HYPERBOLIC FUNCTIONS | SolveeIt

The value of $\frac{e^{2\theta} - 1}{e^{2\theta} + 1}$ is.

$\coth{}\theta$

$\coth{}2\theta$

$\tanh{}\theta$

${\tan h}{}2\theta$

Answer

$\tanh{}\theta$

Explanation

$\frac{e^{2\theta} - 1}{2e^{\theta}} = {\sin h}\theta$

$\frac{e^{2\theta} + 1}{2e^{\theta}} = {\cos h}\theta$

$\frac{e^{2\theta} - 1}{e^{2\theta} + 1} = {\tan h}\theta$ .

Question: The value of \(\frac{e^{2\theta} - 1}{e^{2\theta} + 1}\) is....