In a triangle ABC,if tan left(frac{A-B}{2}rig | vedclass

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(A) Using Napier's Analogy,we have $	an \left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cot \left(\frac{C}{2}ight)$.Given $	an \left(\frac{A-B}{2}

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$2 : 1$

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(A) Using Napier's Analogy,we have $	an \left(\frac{A-B}{2}ight) = \frac{a-b}{a+b} \cot \left(\frac{C}{2}ight)$.Given $	an \left(\frac{A-B}{2}ight) = \frac{1}{3} 	an \left(\frac{A+B}{2}ight)$.Since $A+B+C = 180^{\circ}$,we have $\frac{A+B}{2} = 90^{\circ} - \frac{C}{2}$,so $	an \left(\frac{A+B}{2}ight) = \cot \left(\frac{C}{2}ight)$.Substituting this into the given equation: $\frac{a-b}{a+b} \cot \left(\frac{C}{2}ight) = \frac{1}{3} \cot \left(\frac{C}{2}ight)$.Assuming $\cot \left(\frac{C}{2}ight) 
eq 0$,we get $\frac{a-b}{a+b} = \frac{1}{3}$.Cross-multiplying gives $3(a-b) = a+b$,which simplifies to $3a - 3b = a + b$.Rearranging terms gives $2a = 4b$,or $\frac{a}{b} = \frac{4}{2} = 2$.Thus,$a : b = 2 : 1$.

In a triangle $ABC$,if $\tan \left(\frac{A-B}{2}\right) = \frac{1}{3} \tan \left(\frac{A+B}{2}\right)$,then $a : b =$

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