
What is the value of $\operatorname{sech}\left( 0 \right)$?
Answer
530.4k+ views
Hint: We have to first explain the concept of hyperbolic functions. We use the concept of cosine function and find its inverse to find secant. Then we put the value of $x=0$ in the equation of $\operatorname{sech}\left( x \right)=\dfrac{2}{{{e}^{x}}+{{e}^{-x}}}$.
Complete step-by-step solution:
We use the concept of hyperbolic functions. The hyperbolic functions are analogy of the circular function or the trigonometric functions. The hyperbolic function occurs in the solutions of linear differential equations, calculation of distance and angles in the hyperbolic geometry, Laplace’s equations in the cartesian coordinates.
We know that hyperbolic cosine is the even part of the exponential part.
Therefore, $\cosh \left( x \right)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$.
Hyperbolic secant is the inverse of the function $\cosh \left( x \right)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$.
Therefore, $\operatorname{sech}\left( x \right)=\dfrac{1}{\cosh \left( x \right)}=\dfrac{2}{{{e}^{x}}+{{e}^{-x}}}$.
We now need to find the value of $\operatorname{sech}\left( 0 \right)$.
We put the value of $x=0$.
Therefore, $\operatorname{sech}\left( 0 \right)=\dfrac{1}{\cosh \left( 0 \right)}=\dfrac{2}{{{e}^{0}}+{{e}^{-0}}}=\dfrac{2}{1+1}=\dfrac{2}{2}=1$.
The value of $\operatorname{sech}\left( 0 \right)$ is 1.
Note: Hyperbolic functions are related to the natural exponential function as well the circular sine and cosine functions. They are called “hyperbolic” because the relationship between the sine and cosine functions is the same as a unit hyperbola.
Complete step-by-step solution:
We use the concept of hyperbolic functions. The hyperbolic functions are analogy of the circular function or the trigonometric functions. The hyperbolic function occurs in the solutions of linear differential equations, calculation of distance and angles in the hyperbolic geometry, Laplace’s equations in the cartesian coordinates.
We know that hyperbolic cosine is the even part of the exponential part.
Therefore, $\cosh \left( x \right)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$.
Hyperbolic secant is the inverse of the function $\cosh \left( x \right)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$.
Therefore, $\operatorname{sech}\left( x \right)=\dfrac{1}{\cosh \left( x \right)}=\dfrac{2}{{{e}^{x}}+{{e}^{-x}}}$.
We now need to find the value of $\operatorname{sech}\left( 0 \right)$.
We put the value of $x=0$.
Therefore, $\operatorname{sech}\left( 0 \right)=\dfrac{1}{\cosh \left( 0 \right)}=\dfrac{2}{{{e}^{0}}+{{e}^{-0}}}=\dfrac{2}{1+1}=\dfrac{2}{2}=1$.
The value of $\operatorname{sech}\left( 0 \right)$ is 1.
Note: Hyperbolic functions are related to the natural exponential function as well the circular sine and cosine functions. They are called “hyperbolic” because the relationship between the sine and cosine functions is the same as a unit hyperbola.
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