Question

If x be real, then the expression $\frac{2a(x-1){\sin }^2\theta }{x^2-{\sin }^2\theta }$ does not lie between $2a{\sin }^2\left(\theta /2\right)$ and $2a{\cos }^2\left(\theta /2\right).$

• A. True
• B. False

Answer 1

The correct option is A True

If the given expression by y, then $y(x^2-{\sin }^2\theta )=2a(x-1){\sin }^2\theta$
$\Rightarrow x^{2}y-2ax{\sin }^2\theta (y-2a)=0$
Since x is real $\therefore \Delta \ge 0$ $4a^2{\sin }^4\theta +4y{\sin }^2\theta (y-2a)=0$ Cancel $4{\sin }^2\theta ,a+ivequantity.$ or $y^2-2ay+a^2{\sin }^2\theta =+ive$
$(y-a+a\cos \theta )(y-a-a\cos \theta )=+ive$
$\therefore$ y does not lie between p and q where $p<q$ or y does not lie between $a(1-\cos \theta )$ and $a(1+\cos \theta )$ or $2a{\sin }^2\frac{\theta }{2}$ and $2a{\cos }^2\frac{\theta }{2}.$