If sin (α - β) = $\frac{7}{25}$, sin (α + β) = $\frac{4}{5}$, 0° < α, β <$\frac{π}{4}$, then

find the value of tan2α.

$\frac{4}{3}$

tan(2α) = tan (α + β + α - β)

= $\frac{tan(α+β) + tan(α-β)}{1 - tan(α + β) tan(α - β)}$  ......(i)

sin (α - β) = $\frac{7}{25}$

cos (α+β) = $\frac{4}{5}$

Now put in (i)

tan(2α) = $\frac{\frac{4}{3} + \frac{7}{24}}{1 - \frac{7}{24}  × \frac{3}{4}}$ = $\frac{\frac{24}{25}}{\frac{25}{32}}$

tan(2α) = $\frac{25}{24}$ × $\frac{32}{25}$ = $\frac{4}{3}$