If 0 < A, B < 45, cos (A + B) = \(\frac{24}{25}\) and sin (A - B) = \(\frac{5}{13}\), then find tan 2A.

\(\frac{204}{253}\)

cos (A + B) = \(\frac{24 (B)}{25 (H)}\), therefore, P = 7

sin (A - B) = \(\frac{5(P)}{13(H)}\), therefore, B = 12

tan (A + B) = \(\frac{7}{24}\)

tan (A - B) = \(\frac{5}{12}\)

⇒ tan [x + y] = \(\frac{tan x + tan y}{1 - tan x. tan y}\)

Now,

⇒ tan 2A = tan [ (A + B) + (A - B)] = \(\frac{tan (A + B)\;+\;tan (A - B)}{1 - tan (A + B)\;tan (A - B)}\)

= \(\frac{\frac{7}{24} + \frac{5}{12}}{1 - \frac{7}{24}× \frac{5}{12}}\) = \(\frac{\frac{17}{24}}{1 - \frac{35}{288}}\)  =  \(\frac{\frac{17}{24}}{\frac{253}{288}}\)  =  \(\frac{204}{253}\)