if $$0 < A, B < 45^\circ, \cos(A + B) = \frac{24}{25}$$ and $$\sin(A - B) = \frac{15}{17}$$, then $$\tan 2A$$ is

$$\tan 2A$$

= tan((A + B) + (A - B))

=$$\frac{tan(A + B) + tan(A + B)}{1 - tan(A + B)tan(A + B)}$$ ---(1)

$$(\because tan(a + b) = \frac{tana + tanb}{1 - tana.tanb})$$

tan(A + B) = $$\frac{sin(A + B)}{cos(A - B)}$$

tan(A + B) = $$\frac{\sqrt{1 - cos^2(A + B)}}{cos(A + B)}$$ 

tan(A + B) = $$\frac{\sqrt{1 - (24/25)^2}}{(24/25)}$$

tan(A + B) = $$\frac{\sqrt{49/25}}{(24/25)}$$ = 7/24

tan(A - B) = $$\frac{sin(A - B)}{cos(A - B)}$$

tan(A - B)= $$\frac{sin(A - B)}{\sqrt{1 - sin^2(A - B)}}$$

tan(A - B) = $$\frac{15/17}{\sqrt{1 - (15/17)^2}}$$

tan(A - B) = $$\frac{15/17}{\sqrt{64/17)^2}}$$ = 15/8

From eq(1),

=$$\frac{\frac{7}{24} +\frac{15}{8}}{1 - \frac{7}{24}.\frac{15}{8}}$$

=$$\frac{416}{87}$$