Question

A quadratic equation whose roots are ${\text{cosec}}^2\theta$ and ${\sec }^2\theta$, can be-

• A. $x^2-2x+2=0$
• B. $x^2-3x+3=0$
• C. $x^2-5x+5=0$
• D. $x^2-4x-4=0$

Answer 1

Answer: (C)

$x^2-5x+5=0$

$asse{c}^2\theta \ge 1andcose{c}^2\theta \ge 1henceequationwithroots\ge 1canonlybepossiblesolutionhence,x^2-5x+5=0x=\left(\frac{5\pm \sqrt{25-4\times 5}}{2}\right)=\left(\frac{5\pm \sqrt{5}}{2}\right)asherex\ge 1sothisisonepossibleequationtohavedesiredroots$