In any triangle, prove that - a cosA + b cosB + c cosC​ = r/R
​​                                                         a + b + c

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Please find below the solution to the asked query:

$$\begin{gathered} \text{L.H.S.=}\frac{\mathrm{acosA}+\mathrm{bcosB}+\mathrm{ccosC}}{\mathrm{a}+\mathrm{b}+\mathrm{c}} \\ \mathrm{Use} \mathrm{sine} \mathrm{rule} \\ \frac{\mathrm{a}}{\mathrm{sinA}}=\frac{\mathrm{b}}{\mathrm{sinB}}=\frac{\mathrm{c}}{\mathrm{sinC}}=2\mathrm{R} \\ \Rightarrow \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{2\mathrm{RsinA}.\mathrm{cosA}+2\mathrm{RsinB}.\mathrm{cosB}+2\mathrm{RsinC}.\mathrm{cosC}}{2\mathrm{RsinA}+2\mathrm{RsinB}+2\mathrm{RsinC}} \\ =\frac{2\mathrm{sinA}.\mathrm{cosA}+2\mathrm{sinB}.\mathrm{cosB}+2\mathrm{sinC}.\mathrm{cosC}}{2\left(\mathrm{sinA}+\mathrm{sinB}+\mathrm{sinC}\right)} \\ =\frac{\sin 2\mathrm{A}+\sin 2\mathrm{B}+\sin 2\mathrm{C}}{2\left(\mathrm{sinA}+\mathrm{sinB}+\mathrm{sinC}\right)} \\ \mathrm{Using} \mathrm{conditional} \mathrm{identities} \mathrm{of} \mathrm{trigonometry} \mathrm{we} \mathrm{have} \\ \sin 2\mathrm{A}+\sin 2\mathrm{B}+\sin 2\mathrm{C}=4\mathrm{sinA}.\mathrm{sinB}.\mathrm{sinC} \\ \mathrm{and} \\ \mathrm{sinA}+\mathrm{sinB}+\mathrm{sinC}=4\cos \frac{\mathrm{A}}{2}.\cos \frac{\mathrm{B}}{2}.\cos \frac{\mathrm{C}}{2} \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{4\mathrm{sinA}.\mathrm{sinB}.\mathrm{sinC}}{2\left(4\cos \frac{\mathrm{A}}{2}.\cos \frac{\mathrm{B}}{2}.\cos \frac{\mathrm{C}}{2}\right)} \\ =\frac{\mathrm{sinA}.\mathrm{sinB}.\mathrm{sinC}}{2\cos \frac{\mathrm{A}}{2}.\cos \frac{\mathrm{B}}{2}.\cos \frac{\mathrm{C}}{2}} \\ =\frac{2\sin {\displaystyle \frac{\mathrm{A}}{2}}.\cos {\displaystyle \frac{\mathrm{A}}{2}}.2\sin {\displaystyle \frac{\mathrm{B}}{2}}.\cos {\displaystyle \frac{\mathrm{B}}{2}}.2\sin {\displaystyle \frac{\mathrm{C}}{2}}.\cos {\displaystyle \frac{\mathrm{C}}{2}}}{2\cos \frac{\mathrm{A}}{2}.\cos \frac{\mathrm{B}}{2}.\cos \frac{\mathrm{C}}{2}} \\ =4\sin \frac{\mathrm{A}}{2}.\sin \frac{\mathrm{B}}{2}.\sin \frac{\mathrm{C}}{2} \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \mathrm{r}=4\mathrm{Rsin}\frac{\mathrm{A}}{2}.\sin \frac{\mathrm{B}}{2}.\sin \frac{\mathrm{C}}{2} \\ 4\sin \frac{\mathrm{A}}{2}.\sin \frac{\mathrm{B}}{2}.\sin \frac{\mathrm{C}}{2}=\frac{\mathrm{r}}{\mathrm{R}} \\ \mathrm{Hence} \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{r}}{\mathrm{R}}=\mathrm{R}.\mathrm{H}.\mathrm{S}. \left(\mathrm{Hence} \mathrm{Proved}\right) \end{gathered}$$

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