Question

If $sin\theta +si{n}^2\theta =1$, then find the value of $co{s}^4\theta +co{s}^2\theta -1$.

• A. 2
• B. 1
• C. 0
• D. -1

Answer 1

The correct option is C 0

$sin\theta +si{n}^2\theta =1$
$\therefore 1-si{n}^2\theta =sin\theta$
$co{s}^2\theta =sin\theta$ .......(i) $[\because co{s}^2\theta +si{n}^2\theta =1]$

Thus, $co{s}^4\theta +co{s}^2\theta -1$
$=(co{s}^2\theta {)}^2-(1-co{s}^2\theta )$
$=(sin\theta {)}^2-(sin\theta {)}^2$ [from (i)]
$=0$