$\frac{{1 + {\rm{tanh\;}}\,\,\frac{x}{2}}}{{1 - {\rm{tanh\;}}\,\,\frac{x}{2}}}$ is equal to

JEE MATHEMATICS MCQ: HYPERBOLIC FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

\frac{{1 + \mathrm{tanh\;}\,\,\frac{x}{2}}}{{1 - \mathrm{tanh\;}\,\,\frac{x}{2}}}

is equal to

View answer and explanation

Correct option: B

{e^x}

Option B (

{e^x}

) is correct because simplifying the hyperbolic tangent expression using exponential forms results in

{e^x}

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