If ${sec}^{2} \frac{π}{7}$ and ${tan}^{2} \frac{π}{7}$ are the roots of the equation $a x^{2} + b x + c = 0$ , then the value of $\frac{5 a^{2} - (b^{2} - c^{2})}{(2 a - c)^{2}}$ is

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Solution

Given : The roots of $a x^{2} + b x + c = 0$ be
$α = {sec}^{2} \frac{π}{7}$ and $β = {tan}^{2} \frac{π}{7}$

We know that,
${sec}^{2} \frac{π}{7} - {tan}^{2} \frac{π}{7} = 1 \Rightarrow α - β = 1 \Rightarrow (α - β)^{2} = 1 \Rightarrow \frac{b^{2} - 4 a c}{a^{2}} = 1 \Rightarrow b^{2} = a^{2} + 4 a c \dots (1)$

Now,
$\frac{5 a^{2} - (b^{2} - c^{2})}{(2 a - c)^{2}} = \frac{5 a^{2} - (a^{2} + 4 a c - c^{2})}{(2 a - c)^{2}} = \frac{4 a^{2} - 4 a c + c^{2}}{(2 a - c)^{2}} = \frac{(2 a - c)^{2}}{(2 a - c)^{2}} = 1$

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