If \(\sin \alpha+(\operatorname{Sin} \alpha)^2=1\), then the value of \((\cos \alpha)^{12}+3(\cos \alpha)^{10}+3(\cos \alpha)^8+(\cos \alpha)^6-1\) is

Solution

Given:
\[ \sin \alpha + (\sin \alpha)^2 = 1 \]

Step 1: Simplify the given equation

\[ \sin \alpha = 1 – (\sin \alpha)^2 \]
\[ \sin \alpha = (\cos \alpha)^2 \] (Since \((\sin \alpha)^2 + (\cos \alpha)^2 = 1\))

Step 2: Substitute \(\sin \alpha = (\cos \alpha)^2\) into the expression

\[ (\cos \alpha)^{12} + 3(\cos \alpha)^{10} + 3(\cos \alpha)^8 + (\cos \alpha)^6 – 1 \]
\[ = \left((\cos \alpha)^4 + (\cos \alpha)^2\right)^3 – 1 \]
\[ = \left((\sin \alpha)^2 + (\cos \alpha)^2\right)^3 – 1 \]
\[ = 1^3 – 1 \]
\[ = 1 – 1 \]
\[ = 0 \]

Conclusion

The value of \((\cos \alpha)^{12} + 3(\cos \alpha)^{10} + 3(\cos \alpha)^8 + (\cos \alpha)^6 – 1\) is 0.