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
$a=3,b=3,c=1,d=-1$
$\therefore ac+bd=3\times 1+3\times -1=3-3=0$