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) = \(\frac{25}{24}\) × \(\frac{32}{25}\) = \(\frac{4}{3}\)