Question : If $\sin\theta+\sin^{2}\theta=1$, then the value of $\cos^{12}\theta+3\cos^{10}\theta+3\cos^{8}\theta+\cos^{6}\theta-1$ is:
Option 1: 1
Option 2: 2
Option 3: 3
Option 4: 0
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Correct Answer: 0
Solution : Given: $\sin\theta+\sin^{2}\theta=1$ ⇒ $\sin\theta=1-\sin^{2}\theta$ ⇒ $\sin\theta=\cos^2\theta$ Squaring both sides, we get, ⇒ $\sin^2\theta=\cos^4\theta$ ⇒ $1-\cos^2\theta=\cos^4\theta$ ⇒ $1=\cos^4\theta+\cos^2\theta$ Now, cubing both sides, we get, ⇒ $1^3=(\cos^4\theta+\cos^2\theta)^3$ By using $(a + b)^3 = a^3 + 3a^2b + 3ab^2 +b^3$ ⇒ $1=\cos^{12}\theta+3\cos^{10}\theta+3\cos^{8}\theta+\cos^{6}\theta$ ⇒ $\cos^{12}\theta+3\cos^{10}\theta+3\cos^{8}\theta+\cos^{6}\theta-1=0$ Hence, the correct answer is 0.
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