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Question: The value of \(\sinh \left( {{{\cosh }^{ - 1}}x} \right)\) is A. \(\sqrt {{x^2} + 1} \) B. \(\d...

The value of sinh⁡(cosh⁡−1x)\sinh \left( {{{\cosh }^{ - 1}}x} \right) is
A. x2+1\sqrt {{x^2} + 1}
B. 1x2+1\dfrac{1}{{\sqrt {{x^2} + 1} }}
C. x2−1\sqrt {{x^2} - 1}
D. None of these

Explanation

Solution

We will first substitute the value of cosh⁡−1x{\cosh ^{ - 1}}x as ln⁡(x+x2−1)\ln \left( {x + \sqrt {{x^2} - 1} } \right). And then we will use the identity sinh⁡θ=12(eθ−e−θ)\sinh \theta = \dfrac{1}{2}\left( {{e^\theta } - {e^{ - \theta }}} \right). Then, simplify the expression by using properties of logarithm, eln⁡a=a{e^{\ln a}} = a

Complete step by step solution:
As we know sinh⁡θ\sinh \theta and cosh⁡θ\cosh \theta are hyperbolic functions.
We have to evaluate the value of sinh⁡(cosh⁡−1x)\sinh \left( {{{\cosh }^{ - 1}}x} \right)
The value of cosh⁡−1x{\cosh ^{ - 1}}x is given as ln⁡(x+x2−1)\ln \left( {x + \sqrt {{x^2} - 1} } \right)
Hence, we have to calculate the value of sinh⁡(ln⁡(x+x2−1))\sinh \left( {\ln \left( {x + \sqrt {{x^2} - 1} } \right)} \right)
Also, it is known that sinh⁡θ=12(eθ−e−θ)\sinh \theta = \dfrac{1}{2}\left( {{e^\theta } - {e^{ - \theta }}} \right)
Then the value of sinh⁡(ln⁡(x+x2−1))\sinh \left( {\ln \left( {x + \sqrt {{x^2} - 1} } \right)} \right) is
=12(eln⁡(x+x2−1)−e−ln⁡(x+x2−1)) =12(eln⁡(x+x2−1)−eln⁡(x+x2−1)−1)  = \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 eln⁡a=a{e^{\ln a}} = a
=12(x+x2−1−1x+x2−1)= \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,
=12((x+x2−1)2−1x+x2−1) =12(x2+x2−1+2xx2−1−1x+x2−1) =12(2x2−2+2xx2−1x+x2−1)  = \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
=(x2−1)+xx2−1x+x2−1 =x2−1(x2−1+xx+x2−1) =x2−1  = \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⁡(cosh⁡−1x)\sinh \left( {{{\cosh }^{ - 1}}x} \right) is x2−1\sqrt {{x^2} - 1}

Thus, option C is correct.

Note:
The functions sinh⁡x\sinh x and cosh⁡x\cosh x are hyperbolic functions. And the value of sinh⁡x=ex−e−x2\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2} and cosh⁡=ex+e−x2\cosh = \dfrac{{{e^x} + {e^{ - x}}}}{2}. Here, ex{e^x}is the natural exponential function.