Question

Assertion :For any real value of $\theta \ne (2n+1)\pi$ or $(2n+1)\pi /2$, $n\in I$, the value of the expression $y=\frac{{\cos }^2\theta -1}{{\cos }^2\theta +\cos \theta }$ is $y\le 0$ or $y\ge 2$ (either less than or equal to zero or greater than or equal to two)
Because Reason: $\sec \theta \in (-\infty ,-1]\cup [1,\infty )$ for all real values of $\theta$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

$y=\frac{{\cos }^2\theta -1}{{\cos }^2\theta +\cos \theta }$

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

$=\frac{-{\sin }^2\theta }{2\cos \theta .{\cos }^2\frac{\theta }{2}}$

$=-\frac{4{\sin }^2\frac{\theta }{2}.{\cos }^2\frac{\theta }{2}}{2\cos \theta .{\cos }^2\frac{\theta }{2}}$

$=-\frac{2{\sin }^2\frac{\theta }{2}}{\cos \theta }$

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

$=-[\sec \theta -1]$

$=1-\sec \theta$.
Or
$y=1-\sec \theta$
$\sec \theta =1-y$
Now $\sec \theta ϵ(-\infty ,-1]\cup [1,\infty )$
Or
$\sec \theta <-1$ and $\sec \theta >1$
Or
$1-y<-1$ and $1-y>1$
Or
$y>2$ and $y<0$.