let o be a point inside a triangle ABC such that angle OAC=angle OCB= angle OBA= alpha .then prove that

1)cota +cot b + cot c=cot alpha
2) cosec²a+cosec²b+cosec²c= cosec²alpha

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Please find below the solution to the asked query:



$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \angle \mathrm{OAB}=\angle \mathrm{OBC}=\angle \mathrm{OCA}=\mathrm{\alpha } \\ \mathrm{Now} \\ \mathrm{In} ∆\mathrm{AOC}, \angle \mathrm{CAO}=\mathrm{A}-\angle \mathrm{OAB}=\mathrm{A}-\mathrm{\alpha } \\ \mathrm{Using} \mathrm{this} \mathrm{in} △\mathrm{AOC} \mathrm{we} \mathrm{get} \\ \angle \mathrm{COA}=\mathrm{\pi }-\mathrm{A} \\ \mathrm{In} △\mathrm{COA} \mathrm{Using} \mathrm{sine} \mathrm{rule} \mathrm{we} \mathrm{have} \\ \frac{\sin \left(\mathrm{A}-\mathrm{\alpha }\right)}{\mathrm{OC}}=\frac{\sin \angle \mathrm{COA}}{\mathrm{AC}} \\ \Rightarrow \frac{\sin \left(\mathrm{A}-\mathrm{\alpha }\right)}{\mathrm{OC}}=\frac{\sin \left(\mathrm{\pi }-\mathrm{A}\right)}{\mathrm{b}} \\ \Rightarrow \frac{\sin \left(\mathrm{A}-\mathrm{\alpha }\right)}{\mathrm{OC}}=\frac{\mathrm{sinA}}{\mathrm{b}}...\left(\mathrm{i}\right) \\ \mathrm{Similarly} \mathrm{in} △\mathrm{COB}, \mathrm{suing} \mathrm{sine} \mathrm{rule} \mathrm{you} \mathrm{get} \\ \frac{\sin \left(\mathrm{\alpha }\right)}{\mathrm{OC}}=\frac{\sin \left(\angle \mathrm{COB}\right)}{\mathrm{a}} \\ \Rightarrow \frac{\sin \left(\mathrm{\alpha }\right)}{\mathrm{OC}}=\frac{\sin \left(\mathrm{\pi }-\mathrm{C}\right)}{\mathrm{a}} \\ \Rightarrow \frac{\sin \left(\mathrm{\alpha }\right)}{\mathrm{OC}}=\frac{\sin \left(\mathrm{C}\right)}{\mathrm{a}}...\left(\mathrm{ii}\right) \\ \left(\mathrm{i}\right)/\left(\mathrm{ii}\right) \\ \frac{\sin \left(\mathrm{A}-\mathrm{\alpha }\right)}{\sin \left(\mathrm{\alpha }\right)}=\frac{\mathrm{asinA}}{\mathrm{bsinC}} \\ \mathrm{By} \mathrm{sine} \mathrm{rule} \mathrm{in} △\mathrm{ABC}, \mathrm{a}=\mathrm{ksinA} \mathrm{and} \mathrm{b}=\mathrm{ksinB} \\ \Rightarrow \frac{\sin \left(\mathrm{A}-\mathrm{\alpha }\right)}{\sin \left(\mathrm{\alpha }\right)}=\frac{\mathrm{ksinA}.\mathrm{sinA}}{\mathrm{ksinB}.\mathrm{sinC}} \\ \Rightarrow \frac{\sin \left(\mathrm{A}-\mathrm{\alpha }\right)}{\sin \left(\mathrm{\alpha }\right)}=\frac{\mathrm{sinA}.\sin \left(\mathrm{\pi }-\left(\mathrm{B}+\mathrm{C}\right)\right)}{\mathrm{sinB}.\mathrm{sinC}}=\frac{\mathrm{sinA}.\sin \left(\left(\mathrm{B}+\mathrm{C}\right)\right)}{\mathrm{sinB}.\mathrm{sinC}} \\ \Rightarrow \frac{\mathrm{sinA}.\mathrm{cos\alpha }-\mathrm{cosA}.\mathrm{sin\alpha }}{\mathrm{sin\alpha }}=\frac{\mathrm{sinA}.\left(\mathrm{sinB}.\mathrm{cosC}+\mathrm{sinC}.\mathrm{cosB}\right)}{\mathrm{sinB}.\mathrm{sinC}} \\ \Rightarrow \mathrm{sinA}.\mathrm{sinB}.\mathrm{sinC}.\mathrm{cos\alpha }-\mathrm{sinB}.\mathrm{sinC}.\mathrm{cosA}.\mathrm{sin\alpha } \\ =\mathrm{sinA}.\mathrm{sinB}.\mathrm{cosC}.\mathrm{sin\alpha }+\mathrm{sinA}.\mathrm{sinC}.\mathrm{cosB}.\mathrm{sin\alpha } \\ \mathrm{Divide} \mathrm{through} \mathrm{out} \mathrm{sinA}.\mathrm{sinB}.\mathrm{sinC}.\mathrm{sin\alpha } \\ \Rightarrow \mathrm{cot\alpha }-\mathrm{cotA}=\mathrm{cotC}+\mathrm{cotB} \\ \Rightarrow \mathrm{cotA}+\mathrm{cotB}+\mathrm{cotC}=\mathrm{cot\alpha } \\ \mathrm{Note}: \mathrm{Point} \mathrm{O}\hspace{0.17em}\mathrm{is} \mathrm{known} \mathrm{as} \mathrm{Brocard} \mathrm{point}. \\ \mathrm{Now} \mathrm{try} \mathrm{second} \mathrm{part} \mathrm{yourself} \mathrm{and} \mathrm{get} \mathrm{back} \mathrm{to} \mathrm{us} \mathrm{in} \mathrm{case} \mathrm{you} \mathrm{face} \mathrm{problem}. \end{gathered}$$

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