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?
Answer & explanation
Correct answer: option 2
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) → $\frac{7}{25}$