If $\mathrm{\alpha }$ and $\mathrm{\beta }$ are the roots of the equation, 7x² -3x - 2 = 0,then the value of $\frac{\mathrm{\alpha }}{1-{\mathrm{\alpha }}^2}+\frac{\mathrm{\beta }}{1-{\mathrm{\beta }}^2}$ is equal to

Given quadratic equation 7x² -3x- 2 =0 has roots $\mathrm{\alpha }$ and $\mathrm{\beta }$.
$\mathrm{So}, \mathrm{\alpha }+\mathrm{\beta }=\frac{3}{7} \mathrm{and} \mathrm{\alpha \beta }=-\frac{2}{7}.$
Now,
 $\frac{\mathrm{\alpha }}{1-{\mathrm{\alpha }}^2}+\frac{\mathrm{\beta }}{1-{\mathrm{\beta }}^2}=\frac{\mathrm{\alpha }-{\mathrm{\alpha \beta }}^2+\mathrm{\beta }-{\mathrm{\beta \alpha }}^2}{1-\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2\right)+{\mathrm{\alpha }}^2{\mathrm{\beta }}^2}$
$\begin{array}{l}=\frac{(\mathrm{\alpha }+\mathrm{\beta })-\mathrm{\alpha \beta }(\mathrm{\alpha }+\mathrm{\beta })}{1-\left[(\mathrm{\alpha }+\mathrm{\beta }{)}^2-2\mathrm{\alpha \beta }\right]+(\mathrm{\alpha \beta }{)}^2}\\ =\frac{\frac{3}{7}+\frac{2}{7}\left(\frac{3}{7}\right)}{1-\left[{\left(\frac{3}{7}\right)}^2+\frac{4}{7}\right]+\frac{4}{49}}\\ =\frac{21+6}{49-37+4}=\frac{27}{16}\end{array}$