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Question Number 57469 by olalekan2 last updated on 05/Apr/19

((4tan75)/(1−tan^2 75))=(1/(cos150)) find tan75 in surd form

$$\frac{\mathrm{4}{tan}\mathrm{75}}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{75}}=\frac{\mathrm{1}}{{cos}\mathrm{150}}\: \\ $$$${find}\:{tan}\mathrm{75}\:{in}\:{surd}\:{form} \\ $$

Commented by kaivan.ahmadi last updated on 05/Apr/19

if you want to find tv75 you can solve: tg75=tg(30+45)=((tg30+tg45)/(1−tg30tg45))= ((((√3)/3)+1)/(1−((√3)/3)×1))=((((√3)+3)/3)/((3−(√3))/3))=((3+(√3))/(3−(√3)))=(((3+(√3))^2 )/6)

$${if}\:{you}\:{want}\:{to}\:{find}\:{tv}\mathrm{75} \\ $$$${you}\:{can}\:{solve}: \\ $$$${tg}\mathrm{75}={tg}\left(\mathrm{30}+\mathrm{45}\right)=\frac{{tg}\mathrm{30}+{tg}\mathrm{45}}{\mathrm{1}−{tg}\mathrm{30}{tg}\mathrm{45}}= \\ $$$$\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}+\mathrm{1}}{\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}×\mathrm{1}}=\frac{\frac{\sqrt{\mathrm{3}}+\mathrm{3}}{\mathrm{3}}}{\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{3}}}=\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{3}−\sqrt{\mathrm{3}}}=\frac{\left(\mathrm{3}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{6}} \\ $$

Commented by Kunal12588 last updated on 05/Apr/19

=((3((√3)+1)^2 )/6)=(((1+(√3))^2 )/2)=((4+2(√3))/2)=2+(√3)

$$=\frac{\mathrm{3}\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{6}}=\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$

Commented by olalekan2 last updated on 22/Apr/19

thankz sir

$${thankz}\:{sir} \\ $$

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