Question

If $sinx+si{n}^{2}x=1$, then write the value of $co{s}^{8}x+2co{s}^{6}x+co{s}^{4}x.$

Answer 1

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

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

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

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

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

=1