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1/2(`e^(x)` +`e^(-x)` )+1/2(`e^(x)` -`e^(-x)` /1/2(`e^(x)` +`e^(-x)` )-1/2(`e^(x)`...

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sara212 | Student, Undergraduate | Salutatorian

Posted June 19, 2013 at 6:50 AM via web

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1/2(`e^(x)` +`e^(-x)` )+1/2(`e^(x)` -`e^(-x)` /1/2(`e^(x)` +`e^(-x)` )-1/2(`e^(x)` -`e^(-x)` ) =`e^(2x)`

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aruv | High School Teacher | Valedictorian

Posted June 19, 2013 at 7:16 AM (Answer #2)

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I am answering your question ,though some problem in understanding because of not use of brackets.

You can use following laws of exponents.

`x^m=1/(x^(-m))`

`and`

`x^m xx x^n=x^(m+n)`

`LHS={(1/2)(e^x+e^(x))+(1/2)(e^x-e^(-x))}/{(1/2)(e^x+e^(-x))-(1/2)(e^x-e^(-x))}`

`=((1/2)(e^x+e^(-x)+e^x-e^(-x)))/((1/2)(e^x+e^(-x)-e^x+e^(-x)))`

`=(2e^x)/(2e^(-x))`

`=e^x xx (1/e^(-x))`

`=e^x xx e^x`

`=e^(x+x)`

`=e^(2x)`

`=RHS`

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