Question

If $\sin x+{\sin }^{2}x=1$, then ${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x$ is equal to:

• A. 1
• B. 2
• C. 3
• D. 0

Answer 1

The correct option is A 1

We have $\sin x+{\sin }^{2}x=1\Rightarrow \sin x=1-{\sin }^{2}x={\cos }^{2}x$

$\therefore {\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x$

$={\sin }^{6}x+3{\sin }^{5}x+3{\sin }^{4}x+{\sin }^{3}x$ ......... $[\because \sin x={\cos }^{2}x]$

$={\sin }^{3}x({\sin }^{3}x+3{\sin }^{2}x+3\sin x+1)$

$={\sin }^{3}x(\sin x+1{)}^3$

$=(\sin x+{\sin }^{2}x{)}^3=1$