Question

Prove that LHS = RHS​

Answer 1

Answer:

1

Step-by-step explanation:

$\Rightarrow \mathrm{ \dfrac{1}{1+a^{m-n}} +\dfrac{1}{a+a^{m-n}}}\\\\\\\Rightarrow\mathrm{\dfrac{1}{1+\dfrac{a^m}{a^n}}+\dfrac{1}{1+\dfrac{a^n}{a^m}}}\\\\\\\Rightarrow\mathrm{\dfrac{1}{\dfrac{a^n+a^m}{a^n}} +\dfrac{1}{\dfrac{a^m+a^n}{a^m}}}\\\\\\\Rightarrow\mathrm{\dfrac{a^n}{a^n+a^m} +\dfrac{a^m}{a^n+a^m}}\\\\\\\Rightarrow\mathrm{ \dfrac{a^n+a^m}{a^n+a^m}}\\\\\\\Rightarrow 1$

 

Hence proved.

Answer 2

Take LHS

1/{1+a^(m-n)} +1/{1+a^(n-m)

taking (1+a^(m-n) and (1+a^(n-m) as LCM we get

{1+a^(n-m) +1+a^(m-n)}/{1+a^(n-m)+a^(m-n)+a^(m-m+n-n)}

{2+a^(n-m) +a^(m-n)}/{2+a^(n-m)+a^(m-n}

cancelling out numerator and denominator we get LHS=1

which is equal to RHS

Hence proved

{hope it helps}