Question Number 210989 by mokys last updated on 25/Aug/24
$${prove}\:{tan}\left(\mathrm{72}^{°} \right)={tan}\left(\mathrm{66}^{°} \right)+{tan}\left(\mathrm{36}^{°} \right)+{tan}\left(\mathrm{6}^{°} \right)\:\: \\ $$
Answered by som(math1967) last updated on 26/Aug/24
$${tan}\mathrm{72}={tan}\left(\mathrm{66}+\mathrm{6}\right) \\ $$$$\Rightarrow{tan}\mathrm{72}=\frac{{tan}\mathrm{66}+{tan}\mathrm{6}}{\mathrm{1}−{tan}\mathrm{66}{tan}\mathrm{6}} \\ $$$$\Rightarrow{tan}\mathrm{72}={tan}\mathrm{66}+{tan}\mathrm{6}+{tan}\mathrm{72}{tan}\mathrm{66}{tan}\mathrm{6} \\ $$$$\Rightarrow{tan}\mathrm{72}={tan}\mathrm{66}+{tan}\mathrm{6}+{tan}\mathrm{36}\:\bigstar \\ $$$$\bigstar{tan}\mathrm{66}{tan}\mathrm{6}{tan}\mathrm{72} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{66}{sin}\mathrm{6}{sin}\mathrm{72}}{\mathrm{2}{cos}\mathrm{66}{cos}\mathrm{6}{cos}\mathrm{72}} \\ $$$$=\frac{\left({cos}\mathrm{60}−{cos}\mathrm{72}\right){sin}\mathrm{72}}{\left({cos}\mathrm{60}+{cos}\mathrm{72}\right){cos}\mathrm{72}} \\ $$$$=\frac{{sin}\mathrm{72}−\mathrm{2}{sin}\mathrm{72}{cos}\mathrm{72}}{{cos}\mathrm{72}+\mathrm{2}{cos}^{\mathrm{2}} \mathrm{72}} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{36}{cos}\mathrm{36}−{sin}\mathrm{144}}{\mathrm{2}{cos}^{\mathrm{2}} \mathrm{36}−\mathrm{1}+\mathrm{2}{cos}^{\mathrm{2}} \mathrm{72}} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{36}\left(\mathrm{2}{cos}\mathrm{36}−\mathrm{1}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \mathrm{36}+{cos}\mathrm{144}} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{36}\left(\mathrm{2}{cos}\mathrm{36}−\mathrm{1}\right)}{\mathrm{2}{cos}\mathrm{36}\left(\mathrm{2}{cos}\mathrm{36}−\mathrm{1}\right)} \\ $$$$={tan}\mathrm{36} \\ $$