Question

The expression ${\cos }^4\Theta -{\sin }^4\Theta +2{\sin }^2\Theta$ has the value

• A. 0
• B. ${\sin }^2\Theta$
• C. ${\cos }^2\Theta$
• D. 1

Answer 1

The correct option is D 1

$co{s}^4\theta -si{n}^4\theta +2si{n}^2\theta =(co{s}^2\theta {)}^2-(si{n}^2\theta {)}^2+2si{n}^2\theta$
Now using $a^2-b^2=(a+b)(a-b)$ we have
$(co{s}^2\theta -si{n}^2\theta )(co{s}^2\theta +si{n}^2\theta )+2si{n}^2\theta$
Now $co{s}^2\theta +si{n}^2\theta =1$
Now we have
$1-2si{n}^2\theta +2si{n}^2\theta =1$