Question

If $\sin \theta +{\sin }^2\theta +{\sin }^3\theta =1$, then ${\cos }^6\theta -4{\cos }^4\theta +8{\cos }^2\theta =4$.

• A. True
• B. False

Answer 1

The correct option is A True

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

$⟹\sin \theta (1+{\sin }^2\theta )=1-{\sin }^2\theta$

$⟹\sin \theta (2-{\cos }^2\theta )={\cos }^2\theta$

Squaring on both sides

$⟹{\sin }^2\theta (2-{\cos }^2\theta {)}^2={\cos }^4\theta$

$⟹(1-{\cos }^2\theta )(4+{\cos }^4\theta -4{\cos }^2\theta )={\cos }^4\theta$

$⟹4+{\cos }^4\theta -4{\cos }^2\theta -4{\cos }^2\theta -{\cos }^6\theta +4{\cos }^4\theta ={\cos }^4\theta$

$⟹{\cos }^6\theta -4{\cos }^4\theta +8{\cos }^2\theta =4$

So the relation is $\text{True}$