What is expand 1+((e^x-e^{-x})/2) ?

The solution to expand 1+((e^x-e^{-x})/2) is 1+(e^x)/2-(e^{-x})/2

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expand 1+((e^x-e^{-x})/2)

expand

1 + (\frac{e ^{x} - e ^{- x}}{2})

1 + \frac{e ^{x}}{2} - \frac{e ^{- x}}{2}

Solution steps

1 + (\frac{e ^{x} - e ^{- x}}{2})

Remove parentheses:

(a) = a

= 1 + \frac{e ^{x} - e ^{- x}}{2}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{e ^{x} - e ^{- x}}{2} = \frac{e ^{x}}{2} - \frac{e ^{- x}}{2}

= 1 + \frac{e ^{x}}{2} - \frac{e ^{- x}}{2}

\frac{( 1 - x ^{3} + x ^{2} ) ( 1 - x ^{2} + x ^{3} )}{1 - x ^{3} - x ^{2}}

\frac{5}{( s ^{2} + 1 ) ( s ^{2} + 2 s + 2 )}

\frac{y ( 3 t - sin ( 3 t ) )}{9}

\frac{d ^{2}}{d t ^{2}} ((x^{2}) (2 t - 1))

\frac{j - 1}{( j + 2 ) ( j + 4 )}

What is expand 1+((e^x-e^{-x})/2) ?
The solution to expand 1+((e^x-e^{-x})/2) is 1+(e^x)/2-(e^{-x})/2