Show that 7/2(coth x+1)=7e^x/e^x-e^-x?

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1 Answer
maganbhai P.
Aug 2, 2018

Please see below.

Explanation:

Using definition of Hyperbolic functions :

#cothx=(e^x+e^-x)/(e^x-e^-x)#

Adding #1# , both sides we get

#cothx+1=(e^x+e^-x)/(e^x-e^-x)+1#

#:.cothx+1=(e^x+cancel(e^-x)+e^xcancel(-e^-x))/(e^x-e^-x)#

#:.cothx+1=(e^x+e^x)/(e^x-e^-x)#

#:.cothx+1=(2e^x)/(e^x-e^-x)#

Multiplying both sides by #7/2# ,we get

#7/2(cothx+1)=7/2((2e^x)/(e^x-e^-x))#

#:.7/2(cothx+1)=(7e^x)/(e^x-e^-x))#

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