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Question Number 21266 by oyshi last updated on 18/Sep/17

$$\frac{b^2-c^2}{\tan A}+\frac{c^2-a^2}{\tan B}+\frac{a^2-b^2}{\tan C}=0$$

Answered by myintkhaing last updated on 18/Sep/17

$$\mathrm{By}\mathrm{the}\mathrm{law}\mathrm{of}\mathrm{Sines}$$$$\mathrm{Let}\frac{\mathrm{sinA}}{\mathrm{a}}=\frac{\mathrm{sinB}}{\mathrm{b}}=\frac{\mathrm{sinC}}{\mathrm{c}}=\mathrm{k}$$$$\mathrm{sinA}=\mathrm{ka},\mathrm{sinB}=\mathrm{kb},\mathrm{sinC}=\mathrm{kc}$$$$\mathrm{By}\mathrm{the}\mathrm{law}\mathrm{of}\mathrm{Cosines}$$$$\mathrm{cosA}=\frac{{\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2}{2\mathrm{bc}}$$$$\mathrm{tanA}=\frac{\mathrm{sinA}}{\mathrm{cosA}}=\frac{2\mathrm{kabc}}{{\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2}$$$$\frac{{\mathrm{b}}^2-{\mathrm{c}}^2}{\mathrm{tanA}}=\frac{({\mathrm{b}}^2-{\mathrm{c}}^2)({\mathrm{b}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2)}{2\mathrm{kabc}}=\frac{{\mathrm{b}}^4-{\mathrm{c}}^4-{\mathrm{a}}^2{\mathrm{b}}^2+{\mathrm{c}}^2{\mathrm{a}}^2}{2\mathrm{kabc}}$$$$\mathrm{Similarly},$$$$\frac{{\mathrm{c}}^2-{\mathrm{a}}^2}{\mathrm{tanB}}=\frac{{\mathrm{c}}^4-{\mathrm{a}}^4-{\mathrm{b}}^2{\mathrm{c}}^2+{\mathrm{a}}^2{\mathrm{b}}^2}{2\mathrm{kabc}}\mathrm{and}$$$$\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{\mathrm{tanC}}=\frac{{\mathrm{a}}^4-{\mathrm{b}}^4-{\mathrm{c}}^2{\mathrm{a}}^2+{\mathrm{b}}^2{\mathrm{c}}^2}{2\mathrm{kabc}}$$$$\text{You can't use 'macro parameter character \#' in math mode}$$

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