Question

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

then $\frac{b+c}{a+d}=?$

• A. $1$
• B. $2$
• C. $-2$
• D. $3$

Answer 1

The correct option is D $3$

Given $sin\theta +si{n}^2\theta =1$

$\Rightarrow sin\theta =1-si{n}^2\theta$

$\Rightarrow sin\theta =co{s}^2\theta$

We have $aco{s}^{12}\theta +bco{s}^{10}\theta +cco{s}^8\theta +dco{s}^6\theta -1=0$

$\Rightarrow a(co{s}^2\theta {)}^6+b(co{s}^2\theta {)}^5+c(co{s}^2\theta {)}^4+d(co{s}^2\theta {)}^3-1=0$

$\Rightarrow a(sin\theta {)}^6+b(sin\theta {)}^5+c(sin\theta {)}^4+d(sin\theta {)}^3=1$

$\Rightarrow asi{n}^6\theta +bsi{n}^5\theta +csi{n}^4\theta +dsi{n}^3\theta =1^3$

$\Rightarrow asi{n}^6\theta +bsi{n}^5\theta +csi{n}^4\theta +dsi{n}^3\theta =(sin\theta +si{n}^2\theta {)}^3$

$\Rightarrow asi{n}^6\theta +bsi{n}^5\theta +csi{n}^4\theta +dsi{n}^3\theta =si{n}^3\theta +si{n}^6\theta +3si{n}^4\theta +3si{n}^5\theta$

Comparing the coeeficients of $si{n}^6\theta ,si{n}^5\theta ,si{n}^4\theta ,si{n}^3\theta$ we get

$a=1,b=3,c=3,d=1$

$\Rightarrow \frac{b+c}{a+d}=\frac{3+3}{1+1}=\frac{6}{2}=3$

$\therefore \frac{b+c}{a+d}=3$