If `sintheta+sin^2theta=1,` then prove that `cos^(12)theta+3cos^(10)theta+3cos^8theta+cos^6theta-1=0` Given that `sintheta=1-sin^2theta=1-sin^2theta=cos^2theta`

Explanation

To prove that \( \cos^{12}\theta + 3\cos^{10}\theta + 3\cos^8\theta + \cos^6\theta - 1 = 0 \) given that \( \sin\theta + \sin^2\theta = 1 \), we can follow these steps:

### Step 1: Express \(\sin\theta\) in terms of \(\cos^2\theta\)
From the equation \( \sin\theta + \sin^2\theta = 1 \), we can rearrange it as:
\[ \sin^2\theta = 1 - \sin\theta \]
Let \( x = \sin\theta \). Then we have:
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