Question

If sinx + $si{n}^2$x = 1 then the value of $co{s}^{12}$x + 3$co{s}^{1}0$x + 3$co{s}^8$x + $co{s}^6$x + 1 is equal to

• A. 0
• B. 1
• C. 3
• D. 2

Answer 1

The correct option is B 1

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

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

Now,

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

$\Rightarrow ({\cos }^{4}x{)}^3+3({\cos }^{4}x{)}^2{\cos }^{2}x+3{\cos }^{4}x(co{s}^{2}x{)}^2+({\cos }^{2}x{)}^3$

$\Rightarrow ({\sin }^{2}x{)}^3+3{\sin }^{4}x{\cos }^{2}x+3{\sin }^{2}x{\cos }^{4}x+({\cos }^{2}x{)}^3$

$\Rightarrow [{\sin }^{2}x+{\cos }^{2}x{]}^3$

$\Rightarrow 1^3$

$\Rightarrow 1$.