If sin x + sin^2 x = 1,then the value of the | vedclass

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(A) Given that $\sin x + \sin^2 x = 1$. This implies $\sin x = 1 - \sin^2 x = \cos^2 x$. Let the expression be $E = \cos^{12} x + 3\cos^{10} x + 3\cos

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(A) Given that $\sin x + \sin^2 x = 1$. This implies $\sin x = 1 - \sin^2 x = \cos^2 x$. Let the expression be $E = \cos^{12} x + 3\cos^{10} x + 3\cos^8 x + \cos^6 x - 1$. We can factor the expression as $E = \cos^6 x (\cos^6 x + 3\cos^4 x + 3\cos^2 x + 1) - 1$. Recognizing the binomial expansion $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we have $\cos^6 x + 3\cos^4 x + 3\cos^2 x + 1 = (\cos^2 x + 1)^3$. Thus,$E = \cos^6 x (\cos^2 x + 1)^3 - 1$. Since $\cos^2 x = \sin x$,we substitute this into the expression: $E = (\cos^2 x)^3 (\sin x + 1)^3 - 1 = (\sin x)^3 (\sin x + 1)^3 - 1 = (\sin^2 x + \sin x)^3 - 1$. Given $\sin x + \sin^2 x = 1$,we get $E = (1)^3 - 1 = 1 - 1 = 0$.

If $\sin x + \sin^2 x = 1$,then the value of the expression $\cos^{12} x + 3\cos^{10} x + 3\cos^8 x + \cos^6 x - 1$ is equal to

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