Question

For any real value of $\theta \ne \pi$, the value of the expression $y=\frac{{\cos }^2\theta -1}{{\cos }^2\theta +\cos \theta }is:$

• A. $1\le y\le 2$
• B. $y<0$ and $y>2$
• C. $-1\le y\le 1$
• D. $y\ge 1$

Answer 1

The correct option is B $y<0$ and $y>2$

$co{s}^2\theta -1=(1+cos\theta )(cos\theta -1)$

$y=\frac{(cos\theta +1)(cos\theta -1)}{cos\theta (cos\theta +1)}$

$y=\frac{cos\theta -1}{cos\theta }$

$y=1-secx$

As the range of $secx$ is $\le -1$ and $\ge 1$

Hence $y<0$ and $y>2$