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If $\cos x+\cos ^{2} x=1$ then the value of $\sin ^{12} x+3 \sin ^{10} x+3 \sin ^{8} x+\sin ^{6} x-1$ is equal to
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Given : $\cos x+\cos ^{2} x=1$
$\begin{array}{l}
\Rightarrow \cos x=1-\cos ^{2} x \Rightarrow \cos x=\sin ^{2} x \\
\Rightarrow \cos ^{2} x=\sin ^{4} x \Rightarrow 1-\sin ^{2} x=\sin ^{4} x \\
\Rightarrow \sin ^{4} x+\sin ^{2} x=1
\end{array}$
cubic both sides, we have
$\begin{array}{l}
\sin ^{12} x+\sin ^{6} x+3 \sin ^{6} x\left(\sin ^{4} x+\sin ^{2} x\right)=1 \\
\Rightarrow \sin ^{12} x+\sin ^{6} x+3 \sin ^{10} x+3 \sin ^{8} x=1 \\
\Rightarrow \sin ^{12} x+3 \sin ^{10} x+3 \sin ^{8} x+\sin ^{6} x-1=0
\end{array}$
$\begin{array}{l}
\Rightarrow \cos x=1-\cos ^{2} x \Rightarrow \cos x=\sin ^{2} x \\
\Rightarrow \cos ^{2} x=\sin ^{4} x \Rightarrow 1-\sin ^{2} x=\sin ^{4} x \\
\Rightarrow \sin ^{4} x+\sin ^{2} x=1
\end{array}$
cubic both sides, we have
$\begin{array}{l}
\sin ^{12} x+\sin ^{6} x+3 \sin ^{6} x\left(\sin ^{4} x+\sin ^{2} x\right)=1 \\
\Rightarrow \sin ^{12} x+\sin ^{6} x+3 \sin ^{10} x+3 \sin ^{8} x=1 \\
\Rightarrow \sin ^{12} x+3 \sin ^{10} x+3 \sin ^{8} x+\sin ^{6} x-1=0
\end{array}$
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