Question

If $\sin \theta +{\sin }^2\theta =1$ then the value of ${\cos }^{12}\theta +3{\cos }^{10}\theta +3{\cos }^8\theta +{\cos }^6\theta +2{\cos }^4\theta +2{\cos }^2\theta -2$ is _________

• A. -1
• B. 1
• C. 0
• D. $\frac{1}{2}$

Answer 1

The correct option is B 1

Given $\sin \theta +{\sin }^2\theta =1⟹\sin \theta =1-{\sin }^2\theta ⟹\sin \theta ={\cos }^2\theta \Rightarrow {\sin }^2\theta ={\cos }^4\theta$

${\cos }^{12}\theta +3{\cos }^{10}\theta +3{\cos }^8\theta +{\cos }^6\theta +2{\cos }^4\theta +2{\cos }^2\theta -2$

$⟹({\cos }^4\theta {)}^3+3({\cos }^4\theta {)}^2{\cos }^2\theta +3\left({\cos }^4\theta \right)({\cos }^2\theta {)}^2+({\cos }^2\theta {)}^3+2\left(\right({\cos }^2\theta {)}^2+{\cos }^2\theta -1)$

$⟹({\cos }^4\theta +{\cos }^2\theta {)}^3+2({\sin }^2\theta +{\cos }^2\theta -1)$ $(\because (a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3)$

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