A right triangle is shown below whose sides are 2x, x2 - 1, and x2 + 1.(a)Prove the identity (2x)2 + (x2 - 1)2 = (x2 + 1)2for x > 1.x2 +1.12-1​

Answer:[tex](2x)^2 + (x^2 - 1)^2 = (x^2 + 1)^2[/tex]  PROVEDStep-by-step explanation:The sides of the given right triangle are : [tex]2x , (x^2 -1) , (x^2 +1)[/tex]Now here, To show:  [tex](2x)^2 + (x^2 - 1)^2 = (x^2 + 1)^2[/tex]  .... (1)Now, by ALGEBRAIC IDENTITY: [tex](a \pm b)^2  = a ^2 +b^2 \pm 2ab[/tex]So here, similarly[tex](x^2-1)^2 = (x^2)^2 + (1)^2 - 2 (x^2)\\(x^2+1)^2 = (x^2)^2 + (1)^2+ 2 (x^2)[/tex]Now, substituting the value on Left side of (1) , we get :[tex](2x)^2 + (x^2 - 1)^2   = 4x^2 + (x^2)^2 + (1)^2 - 2 (x^2) = 4x^2 + (x^4) + 1 - 2 x^2  = x^4 + 1 + 2x^2[/tex]  ...... (a)Also, the right side of (1) is:[tex](x^2 + 1)^2 = (x^2)^2 + (1)^2+ 2 (x^2)  = x^4 + 1 + 2x^2[/tex]  ... (b)from(a) and (b) we see that[tex]x^4 + 1 + 2x^2 = x^4 + 1 + 2x^2\\\implies(2x)^2 + (x^2 - 1)^2 = (x^2 + 1)^2[/tex]HENCE PROVED.