Question

$\sin x+{\sin }^{2}x=1,$ then the value of ${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x-1$ is

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

We have,

$\sin x+{\sin }^{2}x=1$

$\sin x=1-{\sin }^{2}x$

We know that ${\sin }^{2}x+{\cos }^{2}x=1$

So $\sin x={\cos }^{2}x$ .....(1)

${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x-1=0$

$({\cos }^{4}x{)}^3+3({\cos }^{4}x{)}^2{\cos }^{2}x+3{\cos }^{4}x({\cos }^{2}x{)}^2+({\cos }^{2}x{)}^3-1=0$ .....(2)

Equation (1) value put in equation (2)

$({\sin }^{2}x{)}^3+3{\sin }^{4}x{\cos }^{2}x+3{\sin }^{2}x{\cos }^{4}x+({\cos }^{2}x{)}^3-1=0$ ......(3)

We know that,

$(a+b{)}^3=a^3+b^3+3ab(a+b)$

Apply this formula in equation $\left(3\right)$

So,

$a=\left({\sin }^{2}x\right)$ $b=\left({\cos }^{2}x\right)$

$({\sin }^{2}x+{\cos }^{2}x{)}^3-1=0$

We know that,

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

$1-1=0$

Hence, this is the answer.