Question

# The value of $\sinh ({\cosh ^{ - 1}}x)$isA. $\sqrt {{x^2} + 1}$B. $\dfrac{1}{{\sqrt {{x^2} + 1} }}$C. $\sqrt {{x^2} - 1}$D. None of those

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.