If sin x = 12/13, where 0° < x < 90°, find the value of 1 - cos\(^2\)x

If sin x = 12/13, where 0° < x < 90°, find the value of 1 - cos\(^2\)x

Answer Details

We know that sin x = opposite/hypotenuse = 12/13, therefore, the adjacent side of the right-angled triangle is √(13^2 - 12^2) = 5. Now, we need to find the value of 1 - cos^2x. Using the identity sin^2x + cos^2x = 1, we can write: cos^2x = 1 - sin^2x Substituting sin x = 12/13, we get: cos^2x = 1 - (12/13)^2 cos^2x = 1 - 144/169 cos^2x = 25/169 Therefore, 1 - cos^2x = 1 - 25/169 = 144/169. Hence, the answer is (d) 144/169.