Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is A

$0$

Explanation for the correct option

Step 1: Given information

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

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

$\Rightarrow \sin x={\cos }^{2}x\mathrm{Or}{\cos }^{2}x=\sin x.......\left(1\right)$

Step 2: According to the question solve the given expression

$\because {\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x–1$

$\Rightarrow \left\{{\left({\cos }^{4}x\right)}^3+3{\left({\cos }^{4}x\right)}^2{\cos }^{2}x+3{\cos }^{4}x{\left({\cos }^{2}x\right)}^2+{\left({\cos }^{2}x\right)}^3–1\right\}$

$\Rightarrow \left\{{\left({\sin }^{2}x\right)}^3+3{\left({\sin }^{2}x\right)}^2{\cos }^{2}x+3{\sin }^{2}x{\left({\cos }^{2}x\right)}^2+{\left({\cos }^{2}x\right)}^3–1\right\}$ (From equation (1))

Using the formula $\left\{\because {(a+b)}^3=a^3+3a^{2}b+3a{b}^2+b^3\right\}$

$\therefore \left\{{({\sin }^{2}x+{\cos }^{2}x)}^3-1\right\}$

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

$$\begin{gathered} \Rightarrow \left\{1-1\right\} \\ \Rightarrow 0 \end{gathered}$$

Hence option (A) is the correct answer.