Question

If roots of $x^2+ax+b=0$ are ${\sec }^2\frac{\pi }{8}$ and ${\csc }^2\frac{\pi }{8}$ then which is the following is correct.

• A. $a-b=0$
• B. $a+b=0$
• C. $2a=b$
• D. $a=2b$

Answer 1

Answer: (B)

$a+b=0$

$x^2+ax+b=0$
Roots are ${\sec }^2\frac{\pi }{8}$ and ${\csc }^2\frac{\pi }{8}$
Sum of the roots =$\frac{-B}{A}=(-a)={\sec }^2\frac{\pi }{8}+{\csc }^2\frac{\pi }{8}$
$-a=\frac{1}{{\cos }^2\frac{\pi }{8}}+\frac{1}{{\sin }^2\frac{\pi }{8}}=\frac{{\sin }^2\frac{z}{8}+{\sin }^2\frac{\pi }{8}}{{\sin }^2\frac{\pi }{8}\times {\cos }^2\frac{\pi }{8}}$
$-a=\frac{1}{{\sin }^2\alpha \times {\cos }^2\alpha }$
$-a=\frac{4}{{\left(2\sin \alpha \cos \alpha \right)}^2}=\frac{4}{{\sin }^{2}2\alpha }=\frac{4}{{\sin }^2\frac{z}{4}}=\frac{4}{{\left(\frac{1}{\sqrt{2}}\right)}^2}$
$-a=8$
$a=-8$
Product of Roots $b={\sec }^2\frac{\pi }{8}\times {\csc }^2\frac{\pi }{8}$
$b=\frac{1}{{\cos }^2\frac{\pi }{8}}\times \frac{1}{{\sin }^2\frac{\pi }{8}}$=$\frac{4}{{\left(2\sin \frac{\pi }{8}\cos \frac{\pi }{8}\right)}^2}$
$b=\frac{4}{{\sin }^2\frac{2\pi }{8}}=\frac{4}{{\sin }^2\frac{\pi }{4}}=\frac{4}{{\left(\frac{1}{\sqrt{2}}\right)}^2}=8$
$b=8,a=-8$
Therefore we can say that $a+b=0$