Question Number 213175 by Davidtim last updated on 31/Oct/24
$${we}\:{can}\:{find}\:{tan}\mathrm{120}\:{by}\:{tan}\left(\mathrm{180}−\mathrm{60}\right) \\ $$$${but}\:{can}\:{not}\:{find}\:{by}\:{tan}\left(\mathrm{90}+\mathrm{30}\right)\:{why}? \\ $$
Answered by efronzo1 last updated on 31/Oct/24
$$\:\mathrm{tan}\:\mathrm{120}°=\:\mathrm{tan}\:\left(\mathrm{90}°+\mathrm{30}°\right)\:=−\:\mathrm{cot}\:\mathrm{30}°=−\sqrt{\mathrm{3}} \\ $$
Answered by MATHEMATICSAM last updated on 31/Oct/24
$$\mathrm{You}\:\mathrm{can}\:\mathrm{find}\:\mathrm{by}\:\mathrm{both} \\ $$$$\mathrm{tan}\left(\mathrm{180}°\:−\:\mathrm{60}°\right)\:=\:−\mathrm{tan60}°\:=\:−\:\sqrt{\mathrm{3}} \\ $$$$\mathrm{tan}\left(\mathrm{90}°\:+\:\mathrm{30}°\right)\:=\:−\:\mathrm{cot30}°\:=\:−\:\sqrt{\mathrm{3}} \\ $$
Answered by MrGaster last updated on 31/Oct/24
$$\mathrm{tan}\left(\mathrm{180}−\mathrm{60}\right)=\mathrm{tan60}=−\sqrt{\mathrm{3}} \\ $$$$\mathrm{tan}\left(\mathrm{90}+\mathrm{30}\right)=\frac{\mathrm{sin}\left(\mathrm{90}+\mathrm{30}\right)}{\mathrm{cos}\left(\mathrm{90}+\mathrm{30}\right)}=\frac{\mathrm{cos30}}{−\mathrm{sin30}}=\frac{\sqrt{\mathrm{3}}/\mathrm{2}}{−\mathrm{1}/\mathrm{2}}=−\sqrt{\mathrm{3}} \\ $$
Answered by a.lgnaoui last updated on 01/Nov/24
$$\mathrm{becsuse}\:\mathrm{cos}\:\mathrm{90}=\mathrm{0}\:\:\:\mathrm{and}\:\boldsymbol{\mathrm{tan}}\:\mathrm{90}\:\mathrm{not}\:\mathrm{determined} \\ $$$$ \\ $$