If Tan(A-B)=724, TanA=43 where A, B are acute then A+B=

tan(A-B)=724 tan A=43tan B=tan(A-(A-B)) =tan A-tan (A-B)1+tan A tan (A-B) =43-7241+43·724 =96-2172+28 =75100=34tan(A+B)=tan A+tan B1-tan A tan B =43+341-43·34 =16+90A+B=π2

If Tan(A-B)=724, TanA=43 where A, B are acute then A+B=

tan(A-B)=724 tan A=43tan B=tan(A-(A-B)) =tan A-tan (A-B)1+tan A tan (A-B) =43-7241+43·724 =96-2172+28 =75100=34tan(A+B)=tan A+tan B1-tan A tan B =43+341-43·34 =16+90A+B=π2

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If $Tan (A - B) = \frac{7}{24}, TanA = \frac{4}{3}$ where $A, B$ are acute then $A + B =$

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$\frac{π}{2}$

$\frac{π}{5}$

$\frac{π}{3}$

$\frac{π}{4}$

answer is D.

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Detailed Solution

$\begin{array}{rcl} \tan (A - B) & = & \frac{7}{24} \tan A = \frac{4}{3} \\ \tan B & = & \tan (A - (A - B)) \\ = & \frac{\tan A - \tan (A - B)}{1 + \tan A \tan (A - B)} \\ = & \frac{\frac{4}{3} - \frac{7}{24}}{1 + \frac{4}{3} \cdot \frac{7}{24}} \\ = & \frac{96 - 21}{72 + 28} \\ = & \frac{75}{100} = \frac{3}{4} \end{array}$

$\begin{array}{rcl} \tan (A + B) & = & \frac{\tan A + \tan B}{1 - \tan A \tan B} \\ = & \frac{\frac{4}{3} + \frac{3}{4}}{1 - \frac{4}{3} \cdot \frac{3}{4}} \\ = & \frac{16 + 9}{0} \\ A + B & = & \frac{π}{2} \end{array}$