Question

If $cosx+co{s}^{2}x=1$ then $si{n}^{12}x+3si{n}^{10}x+3si{n}^{8}x+si{n}^6$ x + 1 =

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is C 2

Given,

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

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

$\therefore \cos x={\sin }^{2}x$

$\Rightarrow {\sin }^{2}x+{\sin }^{4}x=1$

Let $\sin x=a$

$\Rightarrow a^2+a^4=1$

Given,

${\sin }^{12}x+3{\sin }^{10}x+3{\sin }^{8}x+{\sin }^{6}x+1$

$\Rightarrow a^{12}+3a^{10}+3a^8+a^6+1$

$a^{12}+a^{10}+2(a^{10}+a^8)+a^8+a^6+1$

$a^8(a^4+a^2)+2a^6(a^4+a^2)+a^4(a^4+a^2)+1$

$a^8+2a^6+a^4+1$

$a^8+a^6+a^6+a^4+1$

$a^4(a^4+a^2)+a^2(a^4+a^2)+1$

$a^4+a^2+1$

$1+1$

$2$

$\therefore {\sin }^{12}x+3{\sin }^{10}x+3{\sin }^{8}x+{\sin }^{6}x+1=2$