For any theta in left( frac{pi}{4}, frac{pi}{ | vedclass

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(B) Let the expression be $E = 3(\sin 	heta - \cos 	heta)^4 + 6(\sin 	heta + \cos 	heta)^2 + 4\sin^6 	heta$.We know that $(\sin 	heta - \cos 	h

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$13 - 4\cos^6 	heta$

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(B) Let the expression be $E = 3(\sin 	heta - \cos 	heta)^4 + 6(\sin 	heta + \cos 	heta)^2 + 4\sin^6 	heta$.We know that $(\sin 	heta - \cos 	heta)^2 = 1 - \sin 2	heta$ and $(\sin 	heta + \cos 	heta)^2 = 1 + \sin 2	heta$.Thus,$E = 3(1 - \sin 2	heta)^2 + 6(1 + \sin 2	heta) + 4\sin^6 	heta$.Expanding the terms: $E = 3(1 - 2\sin 2	heta + \sin^2 2	heta) + 6 + 6\sin 2	heta + 4\sin^6 	heta$.$E = 3 - 6\sin 2	heta + 3\sin^2 2	heta + 6 + 6\sin 2	heta + 4\sin^6 	heta$.$E = 9 + 3\sin^2 2	heta + 4\sin^6 	heta$.Using $\sin 2	heta = 2\sin 	heta \cos 	heta$,we get $E = 9 + 3(4\sin^2 	heta \cos^2 	heta) + 4\sin^6 	heta = 9 + 12\sin^2 	heta \cos^2 	heta + 4\sin^6 	heta$.Substitute $\sin^2 	heta = 1 - \cos^2 	heta$:$E = 9 + 12(1 - \cos^2 	heta)\cos^2 	heta + 4(1 - \cos^2 	heta)^3$.$E = 9 + 12\cos^2 	heta - 12\cos^4 	heta + 4(1 - 3\cos^2 	heta + 3\cos^4 	heta - \cos^6 	heta)$.$E = 9 + 12\cos^2 	heta - 12\cos^4 	heta + 4 - 12\cos^2 	heta + 12\cos^4 	heta - 4\cos^6 	heta$.$E = 13 - 4\cos^6 	heta$.

For any $\theta \in \left( \frac{\pi}{4}, \frac{\pi}{2} \right)$,the expression $3(\sin \theta - \cos \theta)^4 + 6(\sin \theta + \cos \theta)^2 + 4\sin^6 \theta$ equals

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