Question
Answers

The value of \[\sinh ({\cosh ^{ - 1}}x)\]is
A. \[\sqrt {{x^2} + 1} \]
B. \[\dfrac{1}{{\sqrt {{x^2} + 1} }}\]
C. \[\sqrt {{x^2} - 1} \]
D. None of those

Answer Verified Verified
Hint: Firstly we simplify \[{\cosh ^{ - 1}}x\] and then analyze that the value of \[\sinh x\]. We need to describe \[\sinh x\] and \[\cosh x\] in terms of \[{e^x}\], and on simplification we get our answer.
We, also use the formula of Sridhar Archaya for \[a{x^2} + bx + c = 0\], we have, \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step by step solution: We start with, \[x = \cosh y\]
We use the fact that \[\cos {\text{ }}hy = \dfrac{{{e^y} + {e^{ - y}}}}{2}\]
\[ \Rightarrow x = \dfrac{{{e^y} + {e^{ - y}}}}{2}\]
On simplifying further we get,
\[ \Rightarrow 2x = {e^y} + {e^{ - y}}\]
\[ \Rightarrow 2x = {e^y} + \dfrac{1}{{{e^y}}}\]
\[ \Rightarrow 2x = \dfrac{{{e^{2y}} + 1}}{{{e^y}}}\]
On cross multiplication we get,
\[ \Rightarrow {e^y}2x = {e^{2y}} + 1\]
\[ \Rightarrow {e^{2y}} - 2x{e^y} + 1 = 0\]
Let, \[{e^y} = u\],
\[ \Rightarrow {u^2} - 2xu + 1 = 0\]
By the formula of Sridhar Archaya, we have, for \[a{x^2} + bx + c = 0\], \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[ \Rightarrow u = \dfrac{{ - ( - 2x) \pm \sqrt {{{( - 2x)}^2} - 4.1.1} }}{2}\]
\[ \Rightarrow u = \dfrac{{2x \pm \sqrt {4{x^2} - 4} }}{2}\]
On taking 2 out of the root, we get,
\[ \Rightarrow u = \dfrac{{2x \pm 2\sqrt {{x^2} - 1} }}{2}\]
\[ \Rightarrow u = x \pm \sqrt {{x^2} - 1} \]
On Substituting the value of u, we get,
\[ \Rightarrow {e^y} = x \pm \sqrt {{x^2} - 1} \]
As both of these are positive, we get,
\[ \Rightarrow y = \ln (x \pm \sqrt {{x^2} - 1} )\]
Now, \[\sinh ({\cosh ^{ - 1}}x)\]
\[ = \sinh y\]
On substituting the value of y we get,
\[ = \sinh (\ln (x \pm \sqrt {{x^2} - 1} ))\]
Now as, \[\sinh x = \dfrac{{({e^x} - {e^{ - x}})}}{2}\],
So we get,
\[ = \dfrac{1}{2}({e^{(\ln (x \pm \sqrt {{x^2} - 1} ))}} - {e^{ - (\ln (x \pm \sqrt {{x^2} - 1} ))}})\]
As, \[{e^{\ln (x)}} = x\],
So we have,
\[ = \dfrac{1}{2}(x \pm \sqrt {{x^2} - 1} - \dfrac{1}{{x \pm \sqrt {{x^2} - 1} }})\]
On considering, \[\dfrac{1}{2}(x + \sqrt {{x^2} - 1} - \dfrac{1}{{x + \sqrt {{x^2} - 1} }})\]
On simplifying we get,
\[ = \dfrac{1}{2}(\dfrac{{{{(x + \sqrt {{x^2} - 1} )}^2} - 1}}{{x + \sqrt {{x^2} - 1} }})\]
On applying, \[{{\text{(a + b)}}^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}{\text{ + 2ab}}\], we get,
\[{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{(}}\dfrac{{{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{x}}^{\text{2}}}{\text{ - 1 + 2x}}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} {\text{ - 1}}}}{{{\text{x + }}\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} }}{\text{)}}\]
\[ = \dfrac{1}{2}(\dfrac{{2{x^2} - 2 + 2x\sqrt {{x^2} - 1} }}{{x + \sqrt {{x^2} - 1} }})\]
On taking 2 common, we get,
\[ = \dfrac{2}{2}(\dfrac{{{x^2} - 1 + x\sqrt {{x^2} - 1} }}{{x + \sqrt {{x^2} - 1} }})\]
On taking \[\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} \] common we get,
\[ = \sqrt {{x^2} - 1} (\dfrac{{\sqrt {{x^2} - 1} + x}}{{x + \sqrt {{x^2} - 1} }})\]
On simplifying we get,
\[ = \sqrt {{x^2} - 1} \]
The value of \[\sinh ({\cosh ^{ - 1}}x)\]is, \[\sqrt {{x^2} - 1} \]

Hence the correct option is (C).

Note: You need to remember \[\sinh x = \dfrac{1}{2}({e^x} - {e^{ - x}})\] and \[\cosh x = \dfrac{1}{2}({e^x} + {e^{ - x}})\]. When we find \[{\cosh ^{ - 1}}x\] we take \[y = {\cosh ^{ - 1}}x\] to deal with our given problem. We need to know cos and sin function and \[cosh\] and \[sinh\] function are not the same.