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Reve that \( \sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1 \) |
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Answer» \(sin^6\theta + cos^6\theta + 3sin^2\theta cos^2\theta \) \(= sin^6\theta + cos^6\theta + 3sin^2\theta cos^2\theta(sin^2\theta + cos^2\theta)\) \(= (sin^2\theta + cos^2 \theta )^3\) \((\because (a + b)^3 = a^3 + b^3 + 3ab (a + b))\) \(= 1^3\) \(= 1\) Hence proved. |
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