Given, triangle ABC is right angle.

So, let ∠ B = 90°

Then as per the property of angles in a triangle

∠ A + ∠ B + ∠ C = 180°

As ∠ B = 90°

∠ A + 90° + ∠ C = 180°

Then ∠ A + ∠ C = 180° - 90° = 90°

Now, consider sin ²A + sin ²B + sin ²C

As ∠ B = 90°

sin²A + sin²B + sin²C = sin²A + sin²(90°) + sin²C

= sin²A + 1 + sin²C

From before, we know that ∠ A + ∠ C = 90° ; ∠ C = 90° - ∠ A

sin²A + sin²B + sin²C = sin²A + 1 + sin²( 90° - A)

= sin²A + cos²(A) + 1

[by using the identity cos x = sin ( 90° - x)]

sin²A + sin²B + sin²C = (sin²A + cos²A) + 1

= 1 + 1

= 2

[by using the identity sin²θ + cos²θ = 1]

Therefore, sin²A + sin²B + sin²C = 2.