If $0^\circ < A, B < 45^\circ, \cos(A + B) = \frac{24}{25}$ and $\sin(A - B) = \frac{15}{17}$, then $\tan 2A$ is
Answer & explanation
Correct answer: option 3
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 )
= \(\frac{tan(A+B) + tan(A-B) }{1 - tan(A+B) .tan(A-B) }\)
= \(\frac{7/24+ 15/8 }{1 - 7/24 × 15/8 }\)
= \(\frac{52/24 }{ 1 - 35/64 }\)
= \(\frac{416 }{ 87 }\)