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 equal to

Answers (1)

Given, 

\sin\theta+\sin^{2}\theta=1

\\ \Rightarrow \sin\theta=1- \sin^{2}\theta=\cos^{2}\theta

\cos^{12}\theta+3\cos^{10}\theta+3\cos^{8}\theta+\cos^{6}\theta-1 

= \sin^{6}\theta+3\sin^{5}\theta+3\sin^{4}\theta+\sin^{3}\theta-1

= \sin^{3}\theta[\sin^{3}\theta+3\sin^{2}\theta+3\sin\theta+1]-1

= [(\sin^{2}\theta+\sin\theta)^3]-1

= [(1)^3]-1

= 0

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