Question

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

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• A. 2
• B. 1
• C. 0
• D. -1

Answer 1

The correct option is C

0

$sin\theta +si{n}^2\theta =1$.....(i)

$\therefore 1-si{n}^2\theta =sin\theta$ ....(ii)

Thus, $co{s}^4\theta +co{s}^2\theta -1$

$=(co{s}^2\theta {)}^2-si{n}^2\theta$

$=(1-si{n}^2\theta {)}^2-si{n}^2\theta$

$=(sin\theta {)}^2-si{n}^2\theta$.... from (ii)

$=si{n}^2\theta -si{n}^2\theta$

$=0$