If tan (A-B) = frac{7}{24} and tan A = frac{4 | vedclass

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(A) Given $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B} = \frac{7}{24}$.Substituting $	an A = \frac{4}{3}$,we get $\frac{\frac{4}{3} - 	an

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$\frac{\pi}{2}$

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(A) Given $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B} = \frac{7}{24}$.Substituting $	an A = \frac{4}{3}$,we get $\frac{\frac{4}{3} - 	an B}{1 + \frac{4}{3} 	an B} = \frac{7}{24}$.Cross-multiplying: $24(\frac{4}{3} - 	an B) = 7(1 + \frac{4}{3} 	an B)$.$32 - 24 	an B = 7 + \frac{28}{3} 	an B$.$25 = (24 + \frac{28}{3}) 	an B = \frac{72 + 28}{3} 	an B = \frac{100}{3} 	an B$.$	an B = 25 	imes \frac{3}{100} = \frac{3}{4}$.Now,we need to find $A+B$. Since $	an A = \frac{4}{3}$ and $	an B = \frac{3}{4}$,we have $\cot A = \frac{3}{4}$ and $\cot B = \frac{4}{3}$.Using $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B} = \frac{\frac{4}{3} + \frac{3}{4}}{1 - (\frac{4}{3})(\frac{3}{4})} = \frac{\frac{16+9}{12}}{1-1} = \frac{25/12}{0}$,which is undefined.This implies $A+B = \frac{\pi}{2}$.

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

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