Question

if you know answer both the question together plzzzz.....and thank you ☺​

Answer 1

i) The given equation is:

\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}

=\frac{1}{1+\frac{x^a}{x^b}}+\frac{1}{1+\frac{x^b}{x^a}}

=\frac{1}{\frac{x^b+x^a}{x^b}}+\frac{1}{\frac{x^a+x^b}{x^a}}

=\frac{x^b}{x^b+x^a}+\frac{x^a}{x^b+x^a}

=\frac{x^a+x^b}{x^a+x^b}

=1  = RHS

Hence proved

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ii) LHS:

1/(1 + xᵇ⁻ᵃ + xᶜ⁻ᵃ) + 1/(1 + xᵃ⁻ᵇ + xᶜ⁻ᵇ) + 1/(1 + xᵇ⁻ᶜ + xᵃ⁻ᶜ)

= 1/(1 + xᵇ/xᵃ + xᶜ/xᵃ) + 1/(1 + xᵃ/xᵇ + xᶜ/xᵇ) + 1/(1 + xᵇ/xᶜ + xᵃ/xᶜ)

= 1/{(xᵃ + xᵇ + xᶜ)/xᵃ} + 1/{(xᵇ + xᵃ + xᶜ)/xᵇ} + 1/{(xᶜ + xᵇ + xᵃ)/xᶜ}

= xᵃ/(xᵃ + xᵇ + xᶜ) + xᵇ/(xᵃ + xᵇ + xᶜ) + xᶜ/(xᵃ + xᵇ + xᶜ)

= (xᵃ + xᵇ + xᶜ)/(xᵃ + xᵇ + xᶜ)

= 1 = R.H.S.

Answer 2

1)take LHS

simply, taking {(1+x^(a-b)}{1+x^(b-1)}

now

{1+x^(a-b)+1+x^(b-a)}/{1+x^(a-b)}{1+x^(b-a)}

={2+x^(a-b)+x^(b-a)}/1+x^(b-a)+x^(a-b)+x^(a-b+b-a)

={2+x^(a-b)+x^(b-a)}/(1+x^(b-a)+x^(a-b)+1)

={2+x^(a-b)+x^(b-a)}/{2+x^(b-a)+x^(a-b)

now both numerator and denominator are same so they both will be cancelled out

and we get

=1=RHS

2)simmilary in the second question take the LCM as{1+x^(a-b)+x^(c-b}{1+x^(b-a)+x^(c-a)}{1+x^(b-c)+x^(a-c)}

taking that LCM

we get

{1+x^(a-b)+x(c-b)}{1+x^(b-c)+x^(a-c)}+{1+x^(b-a)+x^(c-a)}{1+x^(b-c)+x^(a-c)}+{1+x^(a-b)+x^(c-b)}{1+x^(b-a)+x^(c-a)}/{1+x^(b-a)+x^(c-a)+x^(a-b)+x^0+x^(c-b)+x^(c-a)}

cancelling out numerator and denominator

we get

LHS=1=RHS

hence proved