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Question Number 8257 by lepan last updated on 04/Oct/16

$$IfA+B+C=90°,showthat$$$$tanAtanB+tanBtanC+tanCtanA=1.$$

Answered by sandy_suhendra last updated on 04/Oct/16

$$=\mathrm{tanA}\mathrm{tanB}+\mathrm{tanC}(\mathrm{tanA}+\mathrm{tanB})$$$$=\mathrm{tanA}\mathrm{tanB}+\mathrm{tanC}\tan (\mathrm{A}+\mathrm{B})[1-\mathrm{tanA}\mathrm{tanB}]$$$$=\mathrm{tanA}\mathrm{tanB}+\mathrm{tanC}\mathrm{cotC}[1-\mathrm{tanA}\mathrm{tanB}]$$$$=\mathrm{tanA}\mathrm{tanB}+1-\mathrm{tanA}\mathrm{tanB}$$$$=1$$$$\mathrm{using}:$$$$\tan (\mathrm{A}+\mathrm{B})=\frac{\mathrm{tanA}+\mathrm{tanB}}{1-\mathrm{tanA}\mathrm{tanB}}\Rightarrow \mathrm{tanA}+\mathrm{tanB}=\tan (\mathrm{A}+\mathrm{B})[1-\mathrm{tanA}\mathrm{tanB}]$$$$\tan (\mathrm{A}+\mathrm{B})=\tan (90°-\mathrm{C})=\mathrm{cotC}$$

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