Question

lf $y=\frac{{\sec }^2\theta -\tan \theta }{{\sec }^2\theta +\tan \theta }$ , then

• A. $y\in (\frac{1}{3},3)$
• B. $y\ne (\frac{1}{3},3)$
• C. $-3<y<-\frac{1}{3}$
• D. $y\in R$

Answer 1

The correct option is A $y\in (\frac{1}{3},3)$

$y=\frac{{sec}^2\theta -tan\theta }{{sec}^2\theta +tan\theta }\Rightarrow y=\frac{1+{tan}^2\theta -tan\theta }{1+{tan}^2\theta +tan\theta }$
Let $tan\theta =x$
Therefore $y=\frac{1+x^2-x}{1+x^2+x}\Rightarrow x^2(y-1)+x(y+1)+y-1=0Asx=tan\theta \therefore xisreal\Rightarrow D\ge 0\Rightarrow {(y+1)}^2-4{(y-1)}^2\ge 0\Rightarrow -3y^2+10y-3\ge 0\Rightarrow 3y^2-10y+3\le 0\Rightarrow yϵ(\frac{1}{3},3)$