Question

(a) To determine the numerical value of given expression.

(b) To determine the numerical value of given expression.

Short answer

(a) The value of \({\mathop{\rm sech}\nolimits} 0 = 1\).

(b) The value of \({\cosh ^{ - 1}}1 = 0\).

Step-by-step solution

Given expressions

(a) The expression is \(\sec h0\).

(b) The expression is \(\cos {h^{ - 1}}1\).

Definition of hyperbolic function

The Definition of hyperbolic function:

\(\begin{array}{c}coshx = \frac{{{e^x} + {e^{ - x}}}}{2}\\sechx = \frac{1}{{coshx}}\end{array}\)

Use the formula and substitute the values

(a)

Obtain the value of the expression \(\sec h0\).

\(\begin{array}{c}{\mathop{\rm sech}\nolimits} x = \frac{1}{{\cosh x}}\\ = \frac{1}{{\frac{{{c^x} + {e^{ - x}}}}{2}}}\\ = \frac{2}{{{e^x} + {e^{ - x}}}}\end{array}\)

Substitute 0 for\(x\):

\(\begin{array}{c}{\mathop{\rm sech}\nolimits} 0 = \frac{2}{{{e^0} + {e^0}}}\\ = \frac{2}{{1 + 1}}\\ = \frac{2}{2}\\ = 1\end{array}\)

Thus, the value of \({\mathop{\rm sech}\nolimits} 0 = 1\).

Use the formula and substitute the values

Obtain the value of the expression, \(\cos {h^{ - 1}}1\).

Use the definition of the inverse hyperbolic function, \({\cosh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} - 1} } \right)\).

Where, \(x \ge 1\).

Substitute 1 for\(x\):

\(\begin{array}{c}{\cosh ^{ - 1}}1 = \ln \left( {1 + \sqrt {{1^2} - 1} } \right)\\ = \ln (1)\\ = 0\end{array}\)

Thus, the value of \({\cosh ^{ - 1}}1 = 0\).