Skip to main contentSkip to solution

In a triangle ABC right angled at B, AB=8 unit and AC=10 unit. What is the value of sin⁡2θ−cos⁡2θ\sin^2\theta - \cos^2\theta where theta is angle ACB ?

Solution

✅ Correct Option: 2

In right triangle ABC with right angle at B: AC is hypotenuse (10), AB is opposite to angle C (8). By Pythagoras: BC=100−64=6BC = \sqrt{100 - 64} = 6. So sin⁡θ=810=45\sin\theta = \frac{8}{10} = \frac{4}{5} and cos⁡θ=610=35\cos\theta = \frac{6}{10} = \frac{3}{5}. Then sin⁡2θ−cos⁡2θ=1625−925=725\sin^2\theta - \cos^2\theta = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question