1.

If `sinx+sin^2x=1`, then find the value of `cos^(12)x +3cos^(10)x + 3 cos^8x + cos^6x-1 `

Answer» Correct Answer - 3
We have `sinx+sin^2x=1`
`or sinx=1-sin^2xorsinx=cos^2x`
Now, `cos^12x+3sin^10x+3sin^4x+sin^3x-2`
`=sin^6x+3sin^5x+3sin^4x-2`
`=(sin^2x)^3+3(sin^2x)^2sinx+3(sin^2x)`
`(sinx)^2+(sinx)^3-2`
`=(sin^2x+sinx)^3-2=(1)^3-2=-1`


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