# NIELIT 2019 Feb Scientist D - Section D: 17

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If $\ Sinx+Sin^{2} x=1$ then $\ Cos^{8}x+ 2 \ Cos^{6} x+ \ Cos^{4} x$ equals to :

1. $0$
2. $-1$
3. $1$
4. $2$

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$\textrm{given that:}$ $\sin x+\sin^2x=1$

$\textrm{above equating can be written as:}$

$\implies$ $\sin x=1-\sin^2x$

$\because$ $\sin^2x+\cos^2x=1$

$\implies$ $\sin x=\cos^2x$...(1)

$\textrm{now we have to find the value of$\cos^8x+2\cos^6x+\cos^4x$}$

$\implies$ $\cos^8x+2\cos^6x+\cos^4x$

$\implies$ $(\cos^4x)^2+2*\cos^4x*\cos^2x+(cos^2x)^2$

$\because$ $\left [ (a+b)^2=a^2+b^2+2ab\right ]$

$\implies$ $(cos^4x+cos^2x)^2$

$\textrm{Form eq(1),$\sin x=\cos^2 x,sin^2x=cos^4x$}$

$\implies$ $(\sin^2x+\sin x)^2$

$\implies$ $1^2$ $...form eq(1)$

$\implies$ $1$

$\textrm{option C is correct.}$
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