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try this...

(e^x + e^(-x))^2 = ... the answer is the radicand of the left side...

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Treat it as a quadratic expression with the power of e^x going from 2, 0 to -2.

You will see after factorizing it.

u must be knowing the formula x^2+y^2+2xy=(x+y)^2.

so (2 + e^2x + e^-2x)^1/2 = (2*e^x*e^-x++ e^2x + e^-2x)^1/2

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

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

Using a common denominator of e^2x within the parenthesis we have:

((2e^2x + e^4x + 1)/e^2x)^1/2

= ((e^4x + 2e^2x +1)/e^2x)^1/2

Factoring within the parenthesis we have:

((e^4x + e^2x + e^2x + 1)/e^2x)^1/2

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

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

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

= e^x + e^-x.

let e^x=cosu+isinu then e^-x=cosu-isinu

e^2x=cos2u+isin2u and e^-2x=cos2u-isin2u

l.h.s.=(2+2cos2u)^1/2=sqrt2(1+cos2u)^1/2=sqrt(2*2cos^2(u))=2cosu

e^x+e^-x=2cosu

so the relation is proved.(i is sqrt(-1))

if x=y

x^2=y^2

square both sides and get rid of the power (1/2)

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

but e^(2x)e^(-2x)=e^(2x-2x)=e^0=1

LHS=RHS