Question

If ${\sec }^2\frac{\pi }{7}$ & ${\tan }^2\frac{\pi }{7}$ are the roots of the equation $a{x}^2+bx+c=0$, then $\frac{5a^2-(b^2-c^2)}{(2a-c{)}^2}$ (wherever defined) is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$1$

We know, ${\sec }^2\frac{\pi }{7}-{\tan }^2\frac{\pi }{7}=1$
$\Rightarrow {({\sec }^2\frac{\pi }{7}+{\tan }^2\frac{\pi }{7})}^2-4{\sec }^2\frac{\pi }{7}{\tan }^2\frac{\pi }{7}=1$
$\Rightarrow \frac{b^2}{a^2}-\frac{4c}{a}=1\Rightarrow b^2-4ac=a^2$
$\Rightarrow 4a^2+c^2-4ac=5a^2-b^2+c^2$
$\Rightarrow (2a-c{)}^2=5a^2-(b^2-c^2)$

$\Rightarrow \frac{5a^2-(b^2-c^2)}{(2a-c{)}^2}=1$