1/(cosh2x-sinh2x) = cosh2x+sinh2x?

Comments

• thesiciliansage

  June 2021

  .. 1/ [ cosh(2x) - sinh(2x) ]

  = 1 ÷ [ ½ (e^(2x) + e^(-2x) ) - ½ (e^(2x) - e^(-2x) ) ]

  = 1 ÷ [ ½ e^(2x) + ½ e^(-2x) - ½ e^(2x) + ½ e^(-2x) ]

  = 1 ÷ [ ½ e^(-2x) + ½ e^(-2x) ]

  = 1 ÷ [ e^(-2x) ]

  = e^(2x)

  = ½ (e^(2x) + e^(-2x) ) + ½ (e^(2x) - e^(-2x) ) ]

  = cosh(2x) + sinh(2x)
• polyhymnio

  June 2021

  Multiply left side by (cosh2x + sinh2x)/(cosh2x + sinh2x)

  (cosh2x + sinh2x)/[(cosh2x + sinh2x)(cosh2x – sinh2x)] =

  (cosh2x + sinh2x)/(cosh²2x – sinh²2x) = cosh2x + sinh2x