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Question Number 128333 by liberty last updated on 06/Jan/21
Prove that tan 30°30′ = (√6) +(√2) −(√3) −2
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{tan}\:\mathrm{30}°\mathrm{30}'\:=\:\sqrt{\mathrm{6}}\:+\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{3}}\:−\mathrm{2} \\ $$
Commented by MJS_new last updated on 06/Jan/21
not true?!  tan (7.5°+180°×n) =−2+(√2)−(√3)+(√6)
$$\mathrm{not}\:\mathrm{true}?! \\ $$$$\mathrm{tan}\:\left(\mathrm{7}.\mathrm{5}°+\mathrm{180}°×{n}\right)\:=−\mathrm{2}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{6}} \\ $$
Commented by liberty last updated on 07/Jan/21
yes. i think it true for tan 7°30′
$$\mathrm{yes}.\:\mathrm{i}\:\mathrm{think}\:\mathrm{it}\:\mathrm{true}\:\mathrm{for}\:\mathrm{tan}\:\mathrm{7}°\mathrm{30}'\: \\ $$

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