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In a $\triangle A B C, \frac{a-b}{a+b}=$
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2323 Upvotes
Verified Answer
The correct answer is:
$\tan \left(\frac{A-B}{2}\right) \tan \frac{C}{2}$
$$
\begin{aligned}
& \frac{a-b}{a+b}=\frac{k \sin A-k \sin B}{k \sin A+k \sin B} apply sine rule
\\
& =\frac{2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)} \\
& =\cot \left(\frac{A+B}{2}\right) \tan \left(\frac{A-B}{2}\right) \\
& =\cot \left(90^{\circ}-\frac{c}{2}\right) \tan \left(\frac{A-B}{2}\right) \\
& =\tan \frac{C}{2} \tan \left(\frac{A-B}{2}\right)
\end{aligned}
$$
\begin{aligned}
& \frac{a-b}{a+b}=\frac{k \sin A-k \sin B}{k \sin A+k \sin B} apply sine rule
\\
& =\frac{2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)}{2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)} \\
& =\cot \left(\frac{A+B}{2}\right) \tan \left(\frac{A-B}{2}\right) \\
& =\cot \left(90^{\circ}-\frac{c}{2}\right) \tan \left(\frac{A-B}{2}\right) \\
& =\tan \frac{C}{2} \tan \left(\frac{A-B}{2}\right)
\end{aligned}
$$
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