​$$\text{If } x=\frac{\sqrt{9}+1}{\sqrt{9}-1} \text{ and } y=\frac{\sqrt{16}+1}{\sqrt{16}-1}, \text{ then the value of } <x^2 - y^2> \text{ is:}$$​​

Correct option is A

Given:

x = $$\frac{\sqrt9+1}{\sqrt9-1}$$

y = $$\frac{\sqrt16+1}{\sqrt16-1}$$  

Solution: 

x²-y²= ?

x = $$\frac{3+1}{3-1}$$ = $$\frac42$$ = 2    

y =  $$\frac{4+1}{4-1}$$ =  $$\frac53$$​    

Now, 

x²-y²= 2^2​ - ($$\frac53$$)² 

= 4 - $$\frac{25}9$$ 

= $$\frac{36-25}9$$

= $$\frac{11}9$$