Given quadratic equation

$7x^2-3x-2=0$

If $\alpha,\beta$ are its roots, then by Vieta's formulas:

$\alpha+\beta=\frac{3}{7}, \qquad \alpha\beta=\frac{-2}{7}$

We need to find

$S=\frac{\alpha}{1-\alpha^2}+\frac{\beta}{1-\beta^2}$

Combine the two terms

Using $1-\alpha^2=(1-\alpha)(1+\alpha), \qquad 1-\beta^2=(1-\beta)(1+\beta),$ we take a common denominator:

$S=\frac{\alpha(1-\beta^2)+\beta(1-\alpha^2)}{(1-\alpha^2)(1-\beta^2)}$

Simplify the numerator

$\alpha(1-\beta^2)+\beta(1-\alpha^2)=\alpha+\beta-\alpha\beta^2-\alpha^2\beta$

Factor the last two terms:

$\alpha\beta^2+\alpha^2\beta=\alpha\beta(\alpha+\beta)$

So numerator becomes

$N=(\alpha+\beta)-\alpha\beta(\alpha+\beta)=(\alpha+\beta)(1-\alpha\beta)$

Substitute values:

$N=\frac{3}{7}\left(1-\left(-\frac{2}{7}\right)\right)=\frac{3}{7}\cdot\frac{9}{7}=\frac{27}{49}$

Simplify the denominator

$D=(1-\alpha^2)(1-\beta^2)=1-(\alpha^2+\beta^2)+\alpha^2\beta^2$

Now,

$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$

$=\left(\frac{3}{7}\right)^2-2\left(-\frac{2}{7}\right)=\frac{9}{49}+\frac{4}{7}=\frac{9}{49}+\frac{28}{49}=\frac{37}{49}$

Also,

$\alpha^2\beta^2=(\alpha\beta)^2=\left(-\frac{2}{7}\right)^2=\frac{4}{49}$

Hence,

$D=1-\frac{37}{49}+\frac{4}{49}=\frac{49-37+4}{49}=\frac{16}{49}$

Compute $S$

$S=\frac{N}{D}=\frac{\frac{27}{49}}{\frac{16}{49}}=\frac{27}{16}$

Match with options

$\frac{27}{16}$ corresponds to Option C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So the answer agrees with the stored answer.