Question

If $\sin \theta +{\sin }^2\theta +{\sin }^3\theta =1$ then the value of ${\cos }^6\theta -4{\cos }^4\theta +8{\cos }^2\theta$ equals

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (C)

$4$

Given $sin\theta +si{n}^2\theta +si{n}^3\theta =1$

$sin\theta +si{n}^3\theta =1-si{n}^2\theta$

$sin\theta \times (1+si{n}^2\theta )=co{s}^2\theta$

Squaring on both sides

$\Rightarrow si{n}^2\theta \times (1+si{n}^2\theta {)}^2=co{s}^4\theta$

$\Rightarrow (1-co{s}^2\theta )\times (1+1-co{s}^2\theta {)}^2=co{s}^4\theta$

$\Rightarrow (1-co{s}^2\theta )\times (2-co{s}^2\theta {)}^2=co{s}^4\theta$

$\Rightarrow (1-co{s}^2\theta )\times (4+co{s}^4\theta -4co{s}^2\theta )=co{s}^4\theta$

$\Rightarrow 4+co{s}^4\theta -4co{s}^2\theta -4co{s}^2\theta -co{s}^6\theta +4co{s}^4\theta =co{s}^4\theta$

$\Rightarrow -co{s}^6\theta +4co{s}^4\theta -8co{s}^2\theta +4=0$

$\Rightarrow co{s}^6\theta -4co{s}^4\theta +8co{s}^2\theta =4$