Question

In triangle ABC, prove the following:

$\frac{a+b}{c}=\frac{\cos \left({\displaystyle \frac{A-B}{2}}\right)}{\sin {\displaystyle \frac{C}{2}}}$

Answer 1

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

Then,
Consider the LHS of the equation $\frac{a+b}{c}=\frac{\cos \left({\displaystyle \frac{A-B}{2}}\right)}{\sin {\displaystyle \frac{C}{2}}}$.
$$\begin{gathered} \mathrm{LHS}=\frac{a+b}{c} \\ =\frac{k\sin A+k\sin B}{k\sin C}\left(\mathrm{using}\left(1\right)\right) \\ =\frac{2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)}{2\sin {\displaystyle \frac{C}{2}}\cos {\displaystyle \frac{C}{2}}} \\ =\frac{\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)}{\sin {\displaystyle \frac{C}{2}}\cos \left({\displaystyle \frac{\mathrm{\pi }-\left(\mathrm{A}+\mathrm{B}\right)}{2}}\right)}\left(\because A+B+C=\mathrm{\pi }\right) \\ =\frac{\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)}{\sin {\displaystyle \frac{C}{2}}\sin \left({\displaystyle \frac{A+B}{2}}\right)} \\ =\frac{\cos \left({\displaystyle \frac{A-B}{2}}\right)}{\sin {\displaystyle \frac{C}{2}}}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$