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

$(a)\;0\qquad(b)\;-1\qquad(c)\;2\qquad(d)\;1$

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1 Answer

$\sin x+\sin^2x=1$
$\Rightarrow \cos^2x+\cos^4x=1$
$(\cos^2x+\cos^4x)^2=1^2$
$(a+b)^2=a^2+2ab+b^2$
$\cos^8x+2\cos^4x\cos^2x+\cos^4x=1$
$\cos^8x+2\cos^6x+\cos^4x=1$
Hence (d) is the correct option.
answered Oct 8, 2013 by sreemathi.v
 
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