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Question: The value of \[\sinh ({\cosh ^{ - 1}}x)\]is A. \[\sqrt {{x^2} + 1} \] B. \[\dfrac{1}{{\sqrt {{x^...

The value of sinh⁡(cosh⁡−1x)\sinh ({\cosh ^{ - 1}}x)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 those

Explanation

Solution

Firstly we simplify cosh⁡−1x{\cosh ^{ - 1}}x and then analyze that the value of sinh⁡x\sinh x. We need to describe sinh⁡x\sinh x and cosh⁡x\cosh x in terms of ex{e^x}, and on simplification we get our answer.
We, also use the formula of Sridhar Archaya for ax2+bx+c=0a{x^2} + bx + c = 0, we have, x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}.

Complete step by step solution: We start with, x=cosh⁡yx = \cosh y
We use the fact that cos⁡ hy=ey+e−y2\cos {\text{ }}hy = \dfrac{{{e^y} + {e^{ - y}}}}{2}
⇒x=ey+e−y2\Rightarrow x = \dfrac{{{e^y} + {e^{ - y}}}}{2}
On simplifying further we get,
⇒2x=ey+e−y\Rightarrow 2x = {e^y} + {e^{ - y}}
⇒2x=ey+1ey\Rightarrow 2x = {e^y} + \dfrac{1}{{{e^y}}}
⇒2x=e2y+1ey\Rightarrow 2x = \dfrac{{{e^{2y}} + 1}}{{{e^y}}}
On cross multiplication we get,
⇒ey2x=e2y+1\Rightarrow {e^y}2x = {e^{2y}} + 1
⇒e2y−2xey+1=0\Rightarrow {e^{2y}} - 2x{e^y} + 1 = 0
Let, ey=u{e^y} = u,
⇒u2−2xu+1=0\Rightarrow {u^2} - 2xu + 1 = 0
By the formula of Sridhar Archaya, we have, for ax2+bx+c=0a{x^2} + bx + c = 0, x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
⇒u=−(−2x)±(−2x)2−4.1.12\Rightarrow u = \dfrac{{ - ( - 2x) \pm \sqrt {{{( - 2x)}^2} - 4.1.1} }}{2}
⇒u=2x±4x2−42\Rightarrow u = \dfrac{{2x \pm \sqrt {4{x^2} - 4} }}{2}
On taking 2 out of the root, we get,
⇒u=2x±2x2−12\Rightarrow u = \dfrac{{2x \pm 2\sqrt {{x^2} - 1} }}{2}
⇒u=x±x2−1\Rightarrow u = x \pm \sqrt {{x^2} - 1}
On Substituting the value of u, we get,
⇒ey=x±x2−1\Rightarrow {e^y} = x \pm \sqrt {{x^2} - 1}
As both of these are positive, we get,
⇒y=ln⁡(x±x2−1)\Rightarrow y = \ln (x \pm \sqrt {{x^2} - 1} )
Now, sinh⁡(cosh⁡−1x)\sinh ({\cosh ^{ - 1}}x)
=sinh⁡y= \sinh y
On substituting the value of y we get,
=sinh⁡(ln⁡(x±x2−1))= \sinh (\ln (x \pm \sqrt {{x^2} - 1} ))
Now as, sinh⁡x=(ex−e−x)2\sinh x = \dfrac{{({e^x} - {e^{ - x}})}}{2},
So we get,
=12(e(ln⁡(x±x2−1))−e−(ln⁡(x±x2−1)))= \dfrac{1}{2}({e^{(\ln (x \pm \sqrt {{x^2} - 1} ))}} - {e^{ - (\ln (x \pm \sqrt {{x^2} - 1} ))}})
As, eln⁡(x)=x{e^{\ln (x)}} = x,
So we have,
=12(x±x2−1−1x±x2−1)= \dfrac{1}{2}(x \pm \sqrt {{x^2} - 1} - \dfrac{1}{{x \pm \sqrt {{x^2} - 1} }})
On considering, 12(x+x2−1−1x+x2−1)\dfrac{1}{2}(x + \sqrt {{x^2} - 1} - \dfrac{1}{{x + \sqrt {{x^2} - 1} }})
On simplifying we get,
=12((x+x2−1)2−1x+x2−1)= \dfrac{1}{2}(\dfrac{{{{(x + \sqrt {{x^2} - 1} )}^2} - 1}}{{x + \sqrt {{x^2} - 1} }})
On applying, (a + b)2 = a2 + b2 + 2ab{{\text{(a + b)}}^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}{\text{ + 2ab}}, we get,
 = 12(x2 + x2 - 1 + 2xx2 - 1 - 1x + x2 - 1){\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{)}}
=12(2x2−2+2xx2−1x+x2−1)= \dfrac{1}{2}(\dfrac{{2{x^2} - 2 + 2x\sqrt {{x^2} - 1} }}{{x + \sqrt {{x^2} - 1} }})
On taking 2 common, we get,
=22(x2−1+xx2−1x+x2−1)= \dfrac{2}{2}(\dfrac{{{x^2} - 1 + x\sqrt {{x^2} - 1} }}{{x + \sqrt {{x^2} - 1} }})
On taking x2 - 1\sqrt {{{\text{x}}^{\text{2}}}{\text{ - 1}}} common we get,
=x2−1(x2−1+xx+x2−1)= \sqrt {{x^2} - 1} (\dfrac{{\sqrt {{x^2} - 1} + x}}{{x + \sqrt {{x^2} - 1} }})
On simplifying we get,
=x2−1= \sqrt {{x^2} - 1}
The value of sinh⁡(cosh⁡−1x)\sinh ({\cosh ^{ - 1}}x)is, x2−1\sqrt {{x^2} - 1}

Hence the correct option is (C).

Note: You need to remember sinh⁡x=12(ex−e−x)\sinh x = \dfrac{1}{2}({e^x} - {e^{ - x}}) and cosh⁡x=12(ex+e−x)\cosh x = \dfrac{1}{2}({e^x} + {e^{ - x}}). When we find cosh⁡−1x{\cosh ^{ - 1}}x we take y=cosh⁡−1xy = {\cosh ^{ - 1}}x to deal with our given problem. We need to know cos and sin function and coshcosh and sinhsinh function are not the same.