# In triangle ABC, prove the following:

Question:

In triangle ABC, prove the following:

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

Solution:

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

We need to prove:

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

Consider

$\mathrm{LHS}=\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}$

$=\frac{\mathrm{ksin} \mathrm{C}}{\mathrm{k}(\sin \mathrm{A}+\sin \mathrm{B})} \quad($ using $(1))$

$=\frac{2 \sin \frac{\mathrm{C}}{2} \cos \frac{\mathrm{C}}{2}}{2 \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right) \cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}$

$=\frac{\sin \frac{\mathrm{C}}{2} \cos \left(\frac{\pi-(\mathrm{A}+\mathrm{B})}{2}\right)}{\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}$ $(\because A+B+C=\pi)$

$=\frac{\sin \frac{\mathrm{C}}{2} \sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)}{\sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)}$

$=\frac{\sin \frac{\mathrm{C}}{2}}{\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)} \ldots(2)$

$\mathrm{RHS}=\frac{1-\tan \frac{A}{2} \tan \frac{B}{2}}{1+\tan \frac{A}{2} \tan \frac{B}{2}}$

$=\frac{1-\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}}{1+\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}}$

$=\frac{\cos \frac{A}{2} \cos \frac{B}{2}-\sin \frac{A}{2} \sin \frac{B}{2}}{\cos \frac{A}{2} \cos \frac{B}{2}+\sin \frac{A}{2} \sin \frac{B}{2}}$

$=\frac{\cos \left(\frac{A+B}{2}\right)}{\cos \left(\frac{A-B}{2}\right)}$

$=\frac{\cos \left(\frac{\pi-\mathrm{C}}{2}\right)}{\cos \left(\frac{A-B}{2}\right)}$ $(\because A+B+C=\pi)$

$=\frac{\sin \frac{\mathrm{C}}{2}}{\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)}=\mathrm{LHS}$   from (2)

Hence proved.