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
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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