Arrange the hyperbolic functions in ascending order $ A = \sinh 0 $ , $ B = \operatorname{Cosh} 0 $ and $ C = \operatorname{Sech} 0 $
A. $ A,B,C $
B. $ A,C,B $
C. $ B,C,A $
D. $ B,A,C $

Hint: The hyperbolic functions have to be written in terms of exponential functions and the value is to be compared at $ x = {0^o} $

Complete step-by-step answer:
The first function is $ A = \sinh 0 $
The formula for $ \sinh x $ in terms of exponential function is,
 $ \sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2} $
Substituting $ x = 0 $ in equation (1),
 $
  \sinh 0 = \dfrac{{{e^0} - {e^{ - 0}}}}{2} \\
  \sinh 0 = \dfrac{{1 - 1}}{2} \\
  \sinh 0 = 0 \\
  $
The value of $ A = 0 \cdots \left( 1 \right) $

The second function is $ B = \cosh 0 $
The formula for in terms of exponential function is,
 $ \cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2} $
Substituting in equation (2),
 $
  \cosh 0 = \dfrac{{{e^0} + {e^{ - 0}}}}{2} \\
  \cosh 0 = \dfrac{{1 + 1}}{2} \\
  \cosh 0 = 1 \\
  $


The value of $ B = 1 \cdots \left( 2 \right) $

The third function is $ C = \operatorname{sech} 0 $
The formula for $ \operatorname{sech} x $ in terms of exponential function is,
 $
  \operatorname{sech} x = \dfrac{1}{{\cosh x}} \\
  \operatorname{sech} x = \dfrac{2}{{{e^x} + {e^{ - x}}}} \\
  $

Substituting $ x = 0 $ in equation (3),
 $
  \operatorname{sech} 0 = \dfrac{2}{{{e^0} + {e^{ - 0}}}} \\
  \operatorname{sech} 0 = \dfrac{2}{2} \\
  \operatorname{sech} 0 = 1 \\
  $
The value of $ C = 1 \cdots \left( 3 \right) $
From equation (1), (2) and (3), the correct increasing order of the hyperbolic functions is
 $ \sinh 0,\cosh 0 = \operatorname{sech} 0 $ Or $ A,B = C $
But this option is not given.
Hence, none of the options is correct.

Note: Hyperbolic functions are very similar to the trigonometric functions but they are expressed in terms of exponential functions. The most common hyperbolic functions are $ \sinh x $ and $ \cosh x $ .
The formula for in terms of exponential function is , $ \cosh x $
 $ \cosh x = \dfrac{{{e^x} + {e^{ - x}}}}{2} $
This function satisfies the $ \cosh 0 = 1 $ and $ \cosh x = \cosh \left( { - x} \right) $ .
The graph of the $ \cosh x $ is always above the graph of $ \dfrac{{{e^x}}}{2} $ and $ \dfrac{{{e^{ - x} \sinh x }}}{2} $
The formula for in terms of exponential function is ,
 $ \sinh x = \dfrac{{{e^x} - {e^{ - x}}}}{2} $
This function satisfies the $ \sinh 0 = 0 $ and $ \sinh \left( { - x} \right) = - \sinh \left( x \right) $ .
The graph of the $ \sinh x $ is always between the graphs of $ \dfrac{{{e^x}}}{2} $ and $ \dfrac{{{e^{ - x}}}}{2} $ .
For the large values of $ x $ the graph of and $ \cosh x $ are closer to each other .

The value of $ \tanh x $ can be calculated by dividing the $ \sinh x $ by $ \cosh x $ as,
 $
  \tanh x = \dfrac{{\sinh x}}{{\cosh x}} \\
  \tanh x = \dfrac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} \\
  $
The factor of $ 2 $ got cancelled in the numerator and denominator.