If sin theta + sin^2 theta = 1 and cos^{12} t | vedclass

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(D) Given $\sin 	heta + \sin^2 	heta = 1$. Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = 1 - \sin^2 	heta = \cos^2 	heta$. Squa

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$ac + bd = 0$

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(D) Given $\sin 	heta + \sin^2 	heta = 1$. Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = 1 - \sin^2 	heta = \cos^2 	heta$. Squaring both sides,$\sin^2 	heta = \cos^4 	heta$,which implies $1 - \cos^2 	heta = \cos^4 	heta$,or $\cos^4 	heta + \cos^2 	heta = 1$. Cubing both sides: $(\cos^4 	heta + \cos^2 	heta)^3 = 1^3$. Expanding using $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$: $(\cos^4 	heta)^3 + 3(\cos^4 	heta)^2(\cos^2 	heta) + 3(\cos^4 	heta)(\cos^2 	heta)^2 + (\cos^2 	heta)^3 = 1$. $\cos^{12} 	heta + 3 \cos^{10} 	heta + 3 \cos^8 	heta + \cos^6 	heta = 1$. $\cos^{12} 	heta + 3 \cos^{10} 	heta + 3 \cos^8 	heta + \cos^6 	heta - 1 = 0$. Comparing this with $\cos^{12} 	heta + a \cos^{10} 	heta + b \cos^8 	heta + c \cos^6 	heta + d = 0$,we get $a = 3, b = 3, c = 1, d = -1$. Thus,$ac + bd = (3)(1) + (3)(-1) = 3 - 3 = 0$.

If $\sin \theta + \sin^2 \theta = 1$ and $\cos^{12} \theta + a \cos^{10} \theta + b \cos^8 \theta + c \cos^6 \theta + d = 0$,then:

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