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Question Number 101082 by 67549972 last updated on 30/Jun/20

$$\mathrm{The}\mathrm{value}\mathrm{of}$$$$e^{{\log }_{10}\tan 1°+{\log }_{10}\tan 2°+{\log }_{10}\tan 3°+...+{\log }_{10}\tan 89°}$$$$\mathrm{is}$$

Commented by Dwaipayan Shikari last updated on 30/Jun/20

$$e^{{log}_{10}(tan1°tan2°tan3°.....tan89°)}$$$$e^{{log}_{10}1}=e^0=1\{tan1°...tan89°=1$$

Commented by smridha last updated on 01/Jul/20

$${\mathit{e}}^{{\mathit{l}\mathit{o}\mathit{g}}_{10}[\mathit{t}\mathit{a}\mathit{n}{1}^{°}.\mathit{t}\mathit{a}\mathit{n}{2}^{°}.\mathit{t}\mathit{a}\mathit{n}{3}^{°}...tan{89}^{°}]}$$$$={\mathit{e}}^{{\mathit{l}\mathit{o}\mathit{g}}_{10}[\mathit{t}\mathit{a}\mathit{n}{1}^{°}.\mathit{t}\mathit{a}\mathit{n}{2}^{°}.\mathit{t}\mathit{a}\mathit{n}3°...\mathit{c}\mathit{o}\mathit{t}{3}^{°}.\mathit{c}\mathit{o}\mathit{t}{2}^{°}.\mathit{c}\mathit{o}\mathit{t}{1}^{°}]}$$$$=e^{{\mathit{l}\mathit{o}\mathit{g}}_{10}1}={\mathit{e}}^0=1.$$

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