What is expand 8/((e^{2x)+e^{-2x})^2} ?

The solution to expand 8/((e^{2x)+e^{-2x})^2} is 8/(e^{4x)+2+e^{-4x}}

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expand 8/((e^{2x)+e^{-2x})^2}

expand

\frac{8}{( e ^{2 x} + e ^{- 2 x} ) ^{2}}

\frac{8}{e ^{4 x} + 2 + e ^{- 4 x}}

Solution steps

\frac{8}{( e ^{2 x} + e ^{- 2 x} ) ^{2}}

(e^{2 x} + e^{- 2 x})^{2} = e^{4 x} + 2 + e^{- 4 x}

= \frac{8}{e ^{4 x} + 2 + e ^{- 4 x}}

p o in t s (42 x - 6 y^{2} - ln ∣ x + y ∣)

\frac{( 4 x - 1 ) ( 2 x ^{2} + x + 2 )}{12}

s (203 sin (θ), 12 + 123 sin (θ))

(2 e^{- x} - 1) (- 2 e^{- y})

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

What is expand 8/((e^{2x)+e^{-2x})^2} ?
The solution to expand 8/((e^{2x)+e^{-2x})^2} is 8/(e^{4x)+2+e^{-4x}}