If tan(A - B) = 7/24 and tanA = 4/3, where A and B are acute, Find the value of (A + B)

We know that, tanA = 4/3

$(tan\left( {A - B} \right) = \frac{{tanA - tanB}}{{1 + tanAtanB}})$

$(\Rightarrow \frac{{\frac{4}{3} - tanB}}{{1 + \frac{4}{3}tanB}} = \frac{7}{{24}})$

$( \Rightarrow \frac{{4 - 3tanB}}{{3 + 4tanB}} = \frac{7}{{24}})$

⇒ 24(4 - 3 tanB) = 7(3 + 4 tanB)

⇒ 96 - 72 tanB = 21 + 28 tanB

⇒ 100 × tanB = 75

⇒ tanB = 75/100 = 3/4

Consider, $(tan\left( {A + B} \right) = \frac{{tanA + tanB}}{{1 - tanAtanB}})$

$(\Rightarrow tan\left( {A + B} \right) = \frac{{\frac{4}{3} + \frac{3}{4}}}{{1 - \frac{4}{3} \times \frac{3}{4}}})$

$(\Rightarrow tan\left( {A + B} \right) = \frac{{\frac{4}{3} + \frac{3}{4}}}{{1 - 1}})$

⇒ tan (A + B) = ∞ (undefined)

∴ A + B = π/2