# 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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Given that: $\sin x+\sin^2x=1\cdots\cdots\to(i)$

Above equation can be written as:

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

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

$\implies \sin x=\cos^2x \cdots\cdots\cdots \to(ii)$

Now we have to find the value of $\cos^8x+2\cos^6x+\cos^4x$

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

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

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

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

From eq. $(i)$, $\sin x=\cos^2 x, \sin^2x = \cos^4x$

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

$\implies 1^2$ $\cdots\cdots \to(1)$

$\implies1$

Option $(C)$ is correct.
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