In a ΔABC right angled at B, AB = 8 unit and AC = 10 unit. What is the value of $\sin^2θ-\cos^2θ$ where θ is ∠ACB?

$\frac{7}{25}$

In a ΔABC right angled at B, AB = 8 unit and AC = 10 unit

AC^2 = AB^2 + BC^2

100 = 64 + BC^2

BC = 6

SinB = 8/10 = 4/5

Sin2B = 16/25

CosB = 6/10 = 3/5

Cos²B = 9/25

Sin²B - Cos²B = 16/25 - 9/25 = 7/25

The correct answer is Option (2) → $\frac{7}{25}$