Question
$\frac{a+b}{a-b}=\tan \frac{A+B}{2} \cot \frac{A-B}{2}$
Step 1
This gives us: \[\tan \frac{A+B}{2} \cot \frac{A-B}{2} = \tan \left(\frac{\pi}{2} - \frac{C}{2}\right) \cot \frac{A-B}{2}\] Show more…
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$(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)=2 c \cot \frac{C}{2}$.
$$ \frac{\cot A+\tan B}{\cot B+\tan A}=\cot A \tan B $$
$\frac{c}{a+b}=\frac{1-\tan \frac{A}{2} \tan \frac{B}{2}}{1+\tan \frac{A}{2} \tan \frac{B}{2}}$
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