Question

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

• A. $0$
• B. $-1$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

Given, $sinx+si{n}^{2}x=1$

$sinx=1-si{n}^{2}x$

$sinx=co{s}^{2}x$

$si{n}^{2}x=co{s}^{4}x$

so,

$co{s}^{8}x+2co{s}^{6}x+co{s}^{4}x$

$=co{s}^{4}x(co{s}^{4}x+2co{s}^{2}x+1)$

$=si{n}^{2}x(si{n}^{2}x+2sinx+1)$

$=si{n}^{4}x+2si{n}^{3}x+si{n}^{2}x$

$=si{n}^{4}x+si{n}^{3}x+si{n}^{3}x+si{n}^{2}x$

$=si{n}^{2}x(si{n}^{2}x+sinx)+sinx(si{n}^{2}x+sinx)$

$=si{n}^{2}x\times \left(1\right)+sinx\left(1\right)$

$=si{n}^{2}x+sinx$

$=1$