1 + tanh x 2 1 - tanh x 2 =, e - x, e x, 2 e x / 2, 2 e - x / 2

1 + tan h x 2 1 - tan h x 2 = 1 + e x 2 - e - x 2 e x 2 + e - x 2 1 - e x 2 - e - x 2 e x 2 + e - x 2 = 2 e x 2 2 e - x 2 = e x

1 + tanh x 2 1 - tanh x 2 =

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Question: Answered & Verified by Expert

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

MathematicsTrigonometric Ratios & IdentitiesAP EAMCETAP EAMCET 2020 (23 Sep Shift 1)

Options:

A $e^{- x}$
B $e^{x}$
C $2 e^{x / 2}$
D $2 e^{- x / 2}$

Solution:

2028 Upvotes Verified Answer

The correct answer is:

e^{x}

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

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

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

$= e^{x}$

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