Question

Choose the value of the hyperbolic trigonometric expression $\dfrac{{\left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]}}$ from the given options:
A. ${e^{ - x}}$
B. ${e^x}$
C. $2{e^{x/2}}$
D. $2{e^{ - x/2}}$

Answer 1

Hint:In this problem, we are using the formulae of hyperbolic trigonometric function that is $\tanh (x) = \dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}$. By using the above-mentioned formula we first find the values of the numerator and denominator of the required hyperbolic trigonometric expression that is $\dfrac{{\left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]}}$ and then divide the numerator and denominator for the required value.

Complete step by step answer:
First, let’s find the value of the numerator.
$Numerator = \left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]$
Now, let’s substitute the value of $\tanh (x/2)$ in the above equation.
$ \Rightarrow Numerator = 1 + \dfrac{{{e^{\dfrac{x}{2}}} - {e^{ - \dfrac{x}{2}}}}}{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}}}}$
On simplification, we get,
$ \Rightarrow Numerator = \dfrac{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}} + {e^{\dfrac{x}{2}}} - {e^{ - \dfrac{x}{2}}}}}{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}}}}$
Here, we can cancel the term ${e^{ - x/2}}$. We get,
$ \Rightarrow Numerator = \dfrac{{2{e^{\dfrac{x}{2}}}}}{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}}}}$

So, the value of the numerator of the required hyperbolic trigonometric expression is $\dfrac{{2{e^{\dfrac{x}{2}}}}}{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}}}}$.
Now, let’s substitute the value of the $\tanh (x/2)$in the above equation.
${\text{Denominator}} = \left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]$
On simplification, we get
$ \Rightarrow denominator = \dfrac{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}} - ({e^{\dfrac{x}{2}}} - {e^{ - \dfrac{x}{2}}})}}{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}}}}$
Here we can eliminate the term ${e^{x/2}}$, we get
$ \Rightarrow denominator = \dfrac{{2{e^{ - \dfrac{x}{2}}}}}{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}}}}$
So, the value of the denominator of the required hyperbolic trigonometric expression is $\dfrac{{2{e^{ - \dfrac{x}{2}}}}}{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}}}}$

Now we require the value of the whole expression.
So, we have, $\dfrac{{\left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]}} = \dfrac{{\dfrac{{2{e^{\dfrac{x}{2}}}}}{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}}}}}}{{\dfrac{{2{e^{ - \dfrac{x}{2}}}}}{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}}}}}}$
Here we can cancel the term ${e^{x/2}} + {e^{ - x/2}}$ in numerator and denominator, so we get
$ \Rightarrow \dfrac{{\left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]}} = \dfrac{{{e^{\dfrac{x}{2}}}}}{{{e^{ - \dfrac{x}{2}}}}}$
On further simplification, we get
$ \Rightarrow \dfrac{{\left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]}} = {e^{\dfrac{x}{2}}} \times {e^{\dfrac{x}{2}}}$
$ \Rightarrow \dfrac{{\left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]}} = {e^{\dfrac{x}{2} + \dfrac{x}{2}}} = {e^x}$
So the required value of the trigonometric expression $\dfrac{{\left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]}}$ is ${e^x}$.

Therefore, the correct option is B.

Note:The formula for hyperbolic trigonometric functions: tangent, sine and cosine is $\dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}$, $\dfrac{{{e^{\dfrac{x}{2}}} - {e^{ - \dfrac{x}{2}}}}}{2}$ and $\dfrac{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}}}}{2}$ respectively. We can also substitute the value of $\tanh (x)$ as $\dfrac{{\sinh x}}{{\cosh x}}$ and then substitute the values of the hyperbolic sine and cosine to get to the required answer.
So, we have, \[\dfrac{{\left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]}} = \dfrac{{\left[ {\dfrac{{\cosh \dfrac{x}{2} + \sinh \dfrac{x}{2}}}{{\cosh \dfrac{x}{2}}}} \right]}}{{\left[ {\dfrac{{\cosh \dfrac{x}{2} - \sinh \dfrac{x}{2}}}{{\cosh \dfrac{x}{2}}}} \right]}}\]
Cancelling the common factors,
\[ \Rightarrow \dfrac{{\left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]}} = \dfrac{{\cosh \dfrac{x}{2} + \sinh \dfrac{x}{2}}}{{\cosh \dfrac{x}{2} - \sinh \dfrac{x}{2}}}\]
Substituting values of hyperbolic sine and cosine, we get,
\[ \Rightarrow \dfrac{{\left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]}} = \dfrac{{\dfrac{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}}}}{2} + \dfrac{{{e^{\dfrac{x}{2}}} - {e^{ - \dfrac{x}{2}}}}}{2}}}{{\dfrac{{{e^{\dfrac{x}{2}}} + {e^{ - \dfrac{x}{2}}}}}{2} - \dfrac{{{e^{\dfrac{x}{2}}} - {e^{ - \dfrac{x}{2}}}}}{2}}}\]
Simplifying the expression, we get,
\[ \Rightarrow \dfrac{{\left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]}} = \dfrac{{{e^{\dfrac{x}{2}}}}}{{{e^{ - \dfrac{x}{2}}}}}\]
\[ \Rightarrow \dfrac{{\left[ {1 + \tanh \left( {\dfrac{x}{2}} \right)} \right]}}{{\left[ {1 - \tanh \left( {\dfrac{x}{2}} \right)} \right]}} = {e^x}\]
Hence, we arrive at the same answer using both ways.