Question

Reve that \( \sin ^{6} \theta+\cos ^{6} \theta+3 \sin ^{2} \theta \cos ^{2} \theta=1 \)

Answer 1

\(sin^6\theta + cos^6\theta + 3sin^2\theta cos^2\theta \)

\(= sin^6\theta + cos^6\theta + 3sin^2\theta cos^2\theta(sin^2\theta + cos^2\theta)\)

\(= (sin^2\theta + cos^2 \theta )^3\)    \((\because (a + b)^3 = a^3 + b^3 + 3ab (a + b))\) 

\(= 1^3\)

\(= 1\)

Hence proved.