2sin2θ+2cos2θ−1=2∗(sin2θ+cos2θ)−1=2−1=12\sin^2\theta + 2\cos^2\theta -1 = 2*(sin^2\theta + \cos^2\theta) - 1 = 2-1 =12sin2θ+2cos2θ−1=2∗(sin2θ+cos2θ)−1=2−1=1
for any value θ\thetaθ
hence the assumption 2sin2θ+2cos2θ−1=02\sin^2\theta + 2\cos^2\theta -1 = 02sin2θ+2cos2θ−1=0 is not true
Answer: The original hypothesis requiring proof is false
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