Question

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Answer 1

Answer:

$\dfrac{a^2+b^2-a^{-2}-b^{-2}}{a^2b^2-a^{-2}b^{-2}}+\dfrac{(a-a^{-1})(b-b^{-1}}{ab+a^{-1}b^{-1}}\\\\\text{Solving the parts one by one :}\\\implies \dfrac{a^2+b^2-(\dfrac{1}{a^2}+\dfrac{1}{b^2})}{a^2b^2-\dfrac{1}{a^2b^2}}}\\\\\implies \dfrac{a^2+b^2-\dfrac{a^2+b^2}{a^2b^2}}{a^2b^2-\dfrac{1}{a^2b^2}}\\\\\implies \dfrac{(a^2+b^2)(1-\dfrac{1}{a^2b^2})}{\dfrac{a^4b^4-1}{a^2b^2}}\\\\\implies \dfrac{(a^2+b^2)(\dfrac{a^2b^2-1}{a^2b^2})}{\dfrac{(a^2b^2+1)(a^2b^2-1)}{a^2b^2}}$

$\implies \dfrac{a^2+b^2}{a^2b^2+1}\rightarrow 1\\\\\text{Next part :}\\\\\implies \dfrac{(a-a^{-1})(b-b^{-1}}{ab+a^{-1}b^{-1}}\\\\\implies \dfrac{(a-\dfrac{1}{a})(b-\dfrac{1}{b})}{ab+\dfrac{1}{ab}}\\\\\implies \dfrac{(\dfrac{a^2-1}{a})(\dfrac{b^2-1}{b})}{\dfrac{a^2b^2+1}{ab}}\\\\\implies \dfrac{(a^2-1)(b^2-1)}{a^2b^2+1}\\\\\implies \dfrac{a^2b^2-a^2-b^2+1}{a^2b^2+1}\rightarrow 2\\\\\text{Adding 1 and 2 we get :}\\\\\implies \dfrac{a^2+b^2}{a^2b^2+1}+\dfrac{a^2b^2+1-a^2-b^2}{a^2b^2+1}$

$\implies \dfrac{a^2+b^2-a^2-b^2+a^2b^2+1}{a^2b^2+1}\\\\\implies \dfrac{a^2b^2+1}{a^2b^2+1}\\\\\implies 1$

The answer is 1 .

Step-by-step explanation:

We are given a problem involving the indices .

The only identity to use is :

$a^{-1}=\dfrac{1}{a}$