1 + tanh x 2 1 - tanh x 2 =

\frac{1 + \tanh (\frac{x}{2})}{1 - \tanh (\frac{x}{2})} =

$e^{- x}$
$e^{x}$
$2 e^{x / 2}$
$2 e^{- x / 2}$

Solution

$\frac{1 + \tan h (\frac{x}{2})}{1 - \tan h (\frac{x}{2})}$

$= \frac{1 + (\frac{e^{\frac{x}{2}} - e^{- \frac{x}{2}}}{e^{\frac{x}{2}} + e^{- \frac{x}{2}}})}{1 - (\frac{e^{\frac{x}{2}} - e^{- \frac{x}{2}}}{e^{\frac{x}{2}} + e^{- \frac{x}{2}}})}$

$= \frac{2 e^{\frac{x}{2}}}{2 e^{- \frac{x}{2}}}$

$= e^{x}$

Asked in: AP EAMCET 2020 (23 Sep Shift 1)

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