AD is the median of triangle ABC and angle BAD = theta ,,and angle CAD=phi ,,prove that cot theta+cot c=cot b+ cot phi

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$$\begin{gathered} \mathrm{This} \mathrm{is} \mathrm{direction} \mathrm{application} \mathrm{m}-\mathrm{n} \mathrm{rule} \mathrm{and} \mathrm{question} \mathrm{is} \mathrm{of} \mathrm{Solution} \mathrm{of} \mathrm{triangle}, \\ \mathrm{not} \mathrm{trigonometric} \mathrm{functions}. \end{gathered}$$


$$\begin{gathered} \mathrm{In} \mathrm{our} \mathrm{question}, \mathrm{AD} \mathrm{is} \mathrm{median}, \mathrm{hence} \\ \frac{\mathrm{AD}}{\mathrm{BD}}=\frac{1}{1}=\frac{\mathrm{m}}{\mathrm{n}} \\ \mathrm{\alpha }=\mathrm{\theta } \mathrm{and} \mathrm{\beta }=\mathrm{\varphi } \\ \mathrm{Applying} \mathrm{m}-\mathrm{n} \mathrm{rule} \mathrm{we} \mathrm{get} \\ \left(1+1\right)\cot \left(\angle \mathrm{ADC}\right)=1 \mathrm{cot\theta }-1 \mathrm{cot\varphi }=1 \mathrm{cotB}-1 \mathrm{cotC} \\ \Rightarrow \mathrm{cot\theta }- \mathrm{cot\varphi }=\mathrm{cotB}-\mathrm{cotC} \\ \Rightarrow \mathrm{cot\theta }+\mathrm{cotC}=\mathrm{cotB}+\mathrm{cot\varphi } \left(\mathrm{Hence} \mathrm{Proved}\right) \end{gathered}$$

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