Solveeit Logo
Login

Question

Question: Find the value of \[\sinh \left( {{{\cosh }^{ - 1}}x} \right)\]. A. \[\sqrt {{x^2} + 1} \] B. \[...

Find the value of sinh⁡(cosh⁡−1x)\sinh \left( {{{\cosh }^{ - 1}}x} \right).
A. x2+1\sqrt {{x^2} + 1}
B. 1x2+1\dfrac{1}{{\sqrt {{x^2} + 1} }}
C. x2−1\sqrt {{x^2} - 1}
D. 1x2−1\dfrac{1}{{\sqrt {{x^2} - 1} }}

Explanation

Solution

Here, we will first use the formula cosh⁡−1x=ln⁡(x+x2−1){\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right) in the given equation and then we will use the formula of hyperbolic sine function, sinh⁡x=ex−e−x2\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2}. Then we will simplify the obtained equation using the properties of trigonometric functions to find the required value.

Complete step by step answer:

We are given that the equation is sinh⁡(cosh⁡−1x)\sinh \left( {{{\cosh }^{ - 1}}x} \right).

Using the formula cosh⁡−1x=ln⁡(x+x2−1){\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right) in the above equation, we get⇒sinh⁡(ln⁡(x+x2−1)) \Rightarrow \sinh \left( {\ln \left( {x + \sqrt {{x^2} - 1} } \right)} \right)

Using the formula of hyperbolic sine function, sinh⁡x=ex−e−x2\sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2} in the above equation, we get
⇒eln⁡(x+x2−1)−eiln⁡(x+x2−1)2\Rightarrow \dfrac{{{e^{\ln \left( {x + \sqrt {{x^2} - 1} } \right)}} - {e^{i\ln \left( {x + \sqrt {{x^2} - 1} } \right)}}}}{2}

Using the value, that is, eln⁡x=x{e^{\ln x}} = x in the above equation, we get

⇒x+x2−1−1x+x2−12 ⇒(x+x2−1)2−1x+x2−12 ⇒(x+x2−1)2−12(x+x2−1)  \Rightarrow \dfrac{{x + \sqrt {{x^2} - 1} - \dfrac{1}{{x + \sqrt {{x^2} - 1} }}}}{2} \\\ \Rightarrow \dfrac{{\dfrac{{{{\left( {x + \sqrt {{x^2} - 1} } \right)}^2} - 1}}{{x + \sqrt {{x^2} - 1} }}}}{2} \\\ \Rightarrow \dfrac{{{{\left( {x + \sqrt {{x^2} - 1} } \right)}^2} - 1}}{{2\left( {x + \sqrt {{x^2} - 1} } \right)}} \\\

Using the formula of trigonometric functions, (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} in the above equation, we get

⇒x2+(x2−1)2−12(x+x2−1) ⇒x2+x2−1+2xx2−1−12(x+x2−1) ⇒2x2−2+2xx2−12(x+x2−1) ⇒2(x2−1+xx2−1)2(x+x2−1) ⇒(x2−1)+xx2−12(x+x2−1)  \Rightarrow \dfrac{{{x^2} + {{\left( {\sqrt {{x^2} - 1} } \right)}^2} - 1}}{{2\left( {x + \sqrt {{x^2} - 1} } \right)}} \\\ \Rightarrow \dfrac{{{x^2} + {x^2} - 1 + 2x\sqrt {{x^2} - 1} - 1}}{{2\left( {x + \sqrt {{x^2} - 1} } \right)}} \\\ \Rightarrow \dfrac{{2{x^2} - 2 + 2x\sqrt {{x^2} - 1} }}{{2\left( {x + \sqrt {{x^2} - 1} } \right)}} \\\ \Rightarrow \dfrac{{2\left( {{x^2} - 1 + x\sqrt {{x^2} - 1} } \right)}}{{2\left( {x + \sqrt {{x^2} - 1} } \right)}} \\\ \Rightarrow \dfrac{{\left( {{x^2} - 1} \right) + x\sqrt {{x^2} - 1} }}{{2\left( {x + \sqrt {{x^2} - 1} } \right)}} \\\

Taking x2−1\sqrt {{x^2} - 1} common from the numerator of the above equation, we get

⇒x2−1(x+x2−1x+x2−1) ⇒x2−1  \Rightarrow \sqrt {{x^2} - 1} \left( {\dfrac{{x + \sqrt {{x^2} - 1} }}{{x + \sqrt {{x^2} - 1} }}} \right) \\\ \Rightarrow \sqrt {{x^2} - 1} \\\

Hence, option D is correct.

Note: In solving these types of questions, the key concept is to have a good understanding of the basic trigonometric values and learn how to use the values from trigonometric tables. Students should have a grasp of trigonometric values, for simplifying the given equation. If we have a root number in the denominator then we will rationalize it. We will divide the numerator and denominator of the fraction by the same number it does not change it. Avoid calculation mistakes.