Answer:
[tex](2x)^2 + (x^2 - 1)^2 = (x^2 + 1)^2[/tex] PROVED
Step-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.
