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

$\frac{416}{87}$

tan2A  =  tan ( A + B + A - B )

ATQ,

cos (A+B) = \(\frac{24}{25}\)

By using pythagoras theorem,

P² + B² = H²

P² + 24² = 25²

P² = 625 - 576 = 49

P = 7

And sin ( A - B ) = \(\frac{15}{17}\)

P² + B² = H²

15² + B² = 17²

B = 8

{ using tan (x + Y ) = \(\frac{tanX + tanY }{1 - tanX.tanY}\) }

Now,

tan2A  =  tan ( A + B + A - B )

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

= \(\frac{7/24+ 15/8 }{1 - 7/24 × 15/8 }\)

= \(\frac{52/24 }{ 1 - 35/64 }\)

= \(\frac{416 }{ 87 }\)