Question

If $\sin x+{\sin }^{2}x=1$, then the value of ${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x-1$ is

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $0$

$\sin x+{\sin }^{2}x=1$
$\Rightarrow {\cos }^{2}x=\sin x$
$\therefore {\cos }^{2}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x-1$
$={\sin }^{6}x+3{\sin }^{5}x+3{\sin }^{4}x+{\sin }^{3}x-1$
$=({\sin }^{2}x+\sin x{)}^3-1=1-1=0$
Hence (a) is correct.