Question

The value of $\sinh \left( {{{\cosh }^{ - 1}}x} \right)$ is
A. $\sqrt {{x^2} + 1} $
B. $\dfrac{1}{{\sqrt {{x^2} + 1} }}$
C. $\sqrt {{x^2} - 1} $
D. None of these

Answer 1

Hint:
We will first substitute the value of ${\cosh ^{ - 1}}x$ as $\ln \left( {x + \sqrt {{x^2} - 1} } \right)$. And then we will use the identity $\sinh \theta = \dfrac{1}{2}\left( {{e^\theta } - {e^{ - \theta }}} \right)$. Then, simplify the expression by using properties of logarithm, ${e^{\ln a}} = a$

Complete step by step solution:
As we know $\sinh \theta $ and $\cosh \theta $ are hyperbolic functions.
We have to evaluate the value of $\sinh \left( {{{\cosh }^{ - 1}}x} \right)$
The value of ${\cosh ^{ - 1}}x$ is given as $\ln \left( {x + \sqrt {{x^2} - 1} } \right)$
Hence, we have to calculate the value of $\sinh \left( {\ln \left( {x + \sqrt {{x^2} - 1} } \right)} \right)$
Also, it is known that $\sinh \theta = \dfrac{1}{2}\left( {{e^\theta } - {e^{ - \theta }}} \right)$
Then the value of $\sinh \left( {\ln \left( {x + \sqrt {{x^2} - 1} } \right)} \right)$ is
$
   = \dfrac{1}{2}\left( {{e^{\ln \left( {x + \sqrt {{x^2} - 1} } \right)}} - {e^{ - \ln \left( {x + \sqrt {{x^2} - 1} } \right)}}} \right) \\
   = \dfrac{1}{2}\left( {{e^{\ln \left( {x + \sqrt {{x^2} - 1} } \right)}} - {e^{\ln {{\left( {x + \sqrt {{x^2} - 1} } \right)}^{ - 1}}}}} \right) \\
$
Next, we will simplify the expression using the property ${e^{\ln a}} = a$
$ = \dfrac{1}{2}\left( {x + \sqrt {{x^2} - 1} - \dfrac{1}{{x + \sqrt {{x^2} - 1} }}} \right)$
On simplifying the above expression, we will get,
$
   = \dfrac{1}{2}\left( {\dfrac{{{{\left( {x + \sqrt {{x^2} - 1} } \right)}^2} - 1}}{{x + \sqrt {{x^2} - 1} }}} \right) \\
   = \dfrac{1}{2}\left( {\dfrac{{{x^2} + {x^2} - 1 + 2x\sqrt {{x^2} - 1} - 1}}{{x + \sqrt {{x^2} - 1} }}} \right) \\
   = \dfrac{1}{2}\left( {\dfrac{{2{x^2} - 2 + 2x\sqrt {{x^2} - 1} }}{{x + \sqrt {{x^2} - 1} }}} \right) \\
$
On taking 2 common
$
   = \dfrac{{\left( {{x^2} - 1} \right) + x\sqrt {{x^2} - 1} }}{{x + \sqrt {{x^2} - 1} }} \\
   = \sqrt {{x^2} - 1} \left( {\dfrac{{\sqrt {{x^2} - 1} + x}}{{x + \sqrt {{x^2} - 1} }}} \right) \\
   = \sqrt {{x^2} - 1} \\
$
Hence, the value of $\sinh \left( {{{\cosh }^{ - 1}}x} \right)$ is \[\sqrt {{x^2} - 1} \]

Thus, option C is correct.

Note:
The functions $\sinh x$ and $\cosh x$ are hyperbolic functions. And the value of $\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}$ and $\cosh = \dfrac{{{e^x} + {e^{ - x}}}}{2}$. Here, ${e^x}$is the natural exponential function.