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Question Number 94523 by peter frank last updated on 19/May/20

Commented by PRITHWISH SEN 2 last updated on 19/May/20

$$\mathrm{By}{\mathrm{Napier}}^{\prime }\mathrm{s}\mathrm{Analogy}$$$$\mathrm{for}\mathrm{any}△\mathrm{ABC}$$$$\mathbf{tan}\frac{\mathbf{B}-\mathbf{C}}{2}=\frac{\mathbf{b}-\mathbf{c}}{\mathbf{b}+\mathbf{c}}\mathbf{cot}\frac{\mathbf{A}}{2}$$$$-\tan (\frac{\mathrm{A}}{2}+\mathrm{B})=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}-\mathrm{c}}.\tan \frac{\mathrm{A}}{2}$$$$\frac{\tan \frac{\mathrm{A}}{2}\mathrm{cotB}+1}{\tan \frac{\mathrm{A}}{2}-\mathrm{cotB}}=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}-\mathrm{c}}\tan \frac{\mathrm{A}}{2}$$$$\tan \frac{\mathrm{A}}{2}\mathrm{cotB}+1=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}-\mathrm{c}}({\tan }^2\frac{\mathrm{A}}{2}-\tan \frac{\mathrm{A}}{2}\cot \mathrm{B})$$$$\tan \frac{\mathrm{A}}{2}\mathrm{cotB}(1+\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}-\mathrm{c}})=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}-\mathrm{c}}{\tan }^2\frac{\mathrm{A}}{2}-1$$$$\frac{2\mathrm{b}}{\mathrm{b}-\mathrm{c}}\mathrm{cotB}=\frac{\mathrm{b}+\mathrm{c}}{\mathrm{b}-\mathrm{c}}\tan \frac{\mathrm{A}}{2}-\cot \frac{\mathrm{A}}{2}$$$$\therefore (\mathbf{b}+\mathbf{c})\mathbf{tan}\frac{\mathbf{A}}{2}-(\mathbf{b}-\mathbf{c})\mathbf{cot}\frac{\mathbf{A}}{2}=2\mathbf{bcotB}\mathbf{Hence}\mathbf{proved}$$

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