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Question Number 70051 by Aditya789 last updated on 30/Sep/19

$$\frac{a+b}{c}=\frac{cos\left(\frac{a-b}{2}\right)}{cos\frac{c}{2}}$$

Answered by $@ty@m123 last updated on 01/Oct/19

$$LHS=\frac{a+b}{c}=\frac{\sin A+\sin B}{\sin C}$$$$=\frac{2\sin \frac{A+B}{2}\cos \frac{A-B}{2}}{2\sin \frac{C}{2}\cos \frac{C}{2}}$$$$=\frac{\sin \frac{\pi -C}{2}\cos \frac{A-B}{2}}{\sin \frac{C}{2}\cos \frac{C}{2}}$$$$=\frac{\cos \frac{C}{2}\cos \frac{A-B}{2}}{\sin \frac{C}{2}\cos \frac{C}{2}}$$$$=\frac{\cos \frac{A-B}{2}}{\sin \frac{C}{2}}$$$$=RHS$$

Answered by behi83417@gmail.com last updated on 30/Sep/19

$$\cos \frac{\mathrm{A}}{2}=\sqrt{\frac{\mathrm{p}(\mathrm{p}-\mathrm{a})}{\mathrm{bc}}},\sin \frac{\mathrm{A}}{2}=\sqrt{\frac{(\mathrm{p}-\mathrm{b})(\mathrm{p}-\mathrm{c})}{\mathrm{bc}}}$$$$\cos \frac{\mathrm{A}-\mathrm{B}}{2}=\sqrt{\frac{\mathrm{p}(\mathrm{p}-\mathrm{a})}{\mathrm{bc}}}.\sqrt{\frac{\mathrm{p}(\mathrm{p}-\mathrm{b})}{\mathrm{ac}}}+$$$$+\sqrt{\frac{(\mathrm{p}-\mathrm{b})(\mathrm{p}-\mathrm{c})}{\mathrm{bc}}}\sqrt{\frac{(\mathrm{p}-\mathrm{a})(\mathrm{p}-\mathrm{c})}{\mathrm{ac}}}=$$$$=\frac{1}{\mathrm{c}}[\mathrm{p}\sqrt{\frac{(\mathrm{p}-\mathrm{a})(\mathrm{p}-\mathrm{b})}{\mathrm{ab}}}+(\mathrm{p}-\mathrm{c})\sqrt{\frac{(\mathrm{p}-\mathrm{a})(\mathrm{p}-\mathrm{b})}{\mathrm{ab}}}]=$$$$=\frac{2\mathrm{p}-\mathrm{c}}{\mathrm{c}}\sqrt{\frac{(\mathrm{p}-\mathrm{a})(\mathrm{p}-\mathrm{b})}{\mathrm{ab}}}=\frac{\mathbf{a}+\mathbf{b}}{\mathbf{c}}.\mathbf{sin}\frac{\mathbf{C}}{2}.$$$$\mathrm{please}\mathrm{check}\mathrm{the}\mathrm{question}\mathrm{sir}.$$

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