Question

In triangle ABC, prove the following:

$\frac{c}{a+b}=\frac{1-\tan \left({\displaystyle \frac{A}{2}}\right)\tan \left({\displaystyle \frac{B}{2}}\right)}{1+\tan \left({\displaystyle \frac{A}{2}}\right)\tan \left({\displaystyle \frac{B}{2}}\right)}$

Answer 1

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$ ...(1)

We need to prove:
$\frac{c}{a+b}=\frac{1-\tan \left({\displaystyle \frac{A}{2}}\right)\tan \left({\displaystyle \frac{B}{2}}\right)}{1+\tan \left({\displaystyle \frac{A}{2}}\right)\tan \left({\displaystyle \frac{B}{2}}\right)}$

Consider

$$\begin{gathered} \mathrm{LHS}=\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}} \\ =\frac{\mathrm{ksinC}}{\mathrm{k}\left(\mathrm{sinA}+\mathrm{sinB}\right)}\left(\mathrm{using}\left(1\right)\right) \\ =\frac{2\sin {\displaystyle \frac{\mathrm{C}}{2}}\cos {\displaystyle \frac{\mathrm{C}}{2}}}{2\sin \left({\displaystyle \frac{\mathrm{A}+\mathrm{B}}{2}}\right)\cos \left({\displaystyle \frac{\mathrm{A}-\mathrm{B}}{2}}\right)} \\ =\frac{\sin {\displaystyle \frac{\mathrm{C}}{2}}\cos \left({\displaystyle \frac{\mathrm{\pi }-\left(\mathrm{A}+\mathrm{B}\right)}{2}}\right)}{\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)}\left(\because A+B+C=\mathrm{\pi }\right) \\ =\frac{\sin {\displaystyle \frac{\mathrm{C}}{2}}\sin \left({\displaystyle \frac{\mathrm{A}+\mathrm{B}}{2}}\right)}{\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)} \\ =\frac{\sin {\displaystyle \frac{\mathrm{C}}{2}}}{\cos \left({\displaystyle \frac{\mathrm{A}-\mathrm{B}}{2}}\right)}...\left(2\right) \\ \mathrm{RHS}=\frac{1-tan{\displaystyle \frac{A}{2}}tan{\displaystyle \frac{B}{2}}}{1+tan{\displaystyle \frac{A}{2}}tan{\displaystyle \frac{B}{2}}} \\ =\frac{1-{\displaystyle \frac{sin{\displaystyle \frac{A}{2}}}{cos{\displaystyle \frac{A}{2}}}}{\displaystyle \frac{sin{\displaystyle \frac{B}{2}}}{cos{\displaystyle \frac{B}{2}}}}}{1+{\displaystyle \frac{sin{\displaystyle \frac{A}{2}}}{cos{\displaystyle \frac{A}{2}}}}{\displaystyle \frac{sin{\displaystyle \frac{B}{2}}}{cos{\displaystyle \frac{B}{2}}}}} \\ =\frac{cos{\displaystyle \frac{A}{2}}cos{\displaystyle \frac{B}{2}}-sin{\displaystyle \frac{A}{2}}sin{\displaystyle \frac{B}{2}}}{cos{\displaystyle \frac{A}{2}}cos{\displaystyle \frac{B}{2}}+sin{\displaystyle \frac{A}{2}}sin{\displaystyle \frac{B}{2}}} \\ =\frac{cos\left({\displaystyle \frac{A+B}{2}}\right)}{cos\left({\displaystyle \frac{A-B}{2}}\right)} \\ =\frac{cos\left({\displaystyle \frac{\mathrm{\pi }-\mathrm{C}}{2}}\right)}{cos\left({\displaystyle \frac{A-B}{2}}\right)}\left(\because A+B+C=\mathrm{\pi }\right) \\ =\frac{\sin {\displaystyle \frac{\mathrm{C}}{2}}}{\cos \left({\displaystyle \frac{\mathrm{A}-\mathrm{B}}{2}}\right)}=\mathrm{LHS}\left(\mathrm{from}\left(2\right)\right) \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$