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In a triangle ABC right angled at B, AB=8 unit and AC=10 unit. What is the value of sin2θcos2θ\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=10064=6BC = \sqrt{100 - 64} = 6. So sinθ=810=45\sin\theta = \frac{8}{10} = \frac{4}{5} and cosθ=610=35\cos\theta = \frac{6}{10} = \frac{3}{5}. Then sin2θcos2θ=1625925=725\sin^2\theta - \cos^2\theta = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}.

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