Question

If $\sin \theta+\sin ^{2} \theta=1$ and $\cos ^{12} \theta +a \cos ^{10} \theta +b \cos ^{8} \theta +c \cos ^{6} \theta +d=0$, then

• A. ab = cd
• B. ac = bd
• C. ab + cd = 0
• D. ac + bd = 0

Answer 1

Answer: (D)

We have,
$\sin \theta+\sin ^{2} \theta =1$
$\Rightarrow \sin \theta=1-\sin ^{2} \theta$
$\Rightarrow \sin \theta=\cos ^{2} \theta$
We know that,
$\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{3}-1=0$
$\Rightarrow \left(\cos ^{4} \theta+\cos ^{2} \theta\right)^{3}-1=0$
$\Rightarrow \left(\cos ^{12} \theta+3 \cos ^{10} \theta+3 \cos ^{8} \theta+\cos ^{6} \theta\right)-1=0$
$\Rightarrow \cos ^{12} \theta+3 \cos ^{10} \theta+3 \cos ^{8} \theta+\cos ^{6} \theta-1=0$
On comparing, we get
$a=3, b=3, c=1, d=-1$
$\therefore a c +b d=3 \times 1+3 \times-1=3-3=0$