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Question

If $$\sin x+\sin^{2}x=1$$ then $$\cos^{8}x+2 \cos^{6}x+\cos^{4}x$$ is equal to


A
0
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B
1
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C
2
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D
1
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Solution

The correct option is D $$1$$
Given, $$sinx +sin^2x=1$$

$$sinx=1-sin^2x\\$$
$$sinx=cos^2x\\$$
$$sin^2x=cos^4x\\$$

so,
$$cos^8x+2cos^6x+cos^4x\\$$
$$=cos^4x(cos^4x+2cos^2x+1)\\$$
$$=sin^2x (sin^2x+2sinx+1)\\$$
$$=sin^4x+2sin^3x+sin^2x\\$$
$$=sin^4x+sin^3x+sin^3x+sin^2x\\$$
$$=sin^2x(sin^2x+sinx)+ sinx(sin^2x+sinx)\\$$
$$=sin^2x×(1)+ sinx(1)\\$$
$$=sin^2x+sinx\\$$
$$= 1$$

Mathematics

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